A quadratic equation is a polynomial equation of degree 2, which means it contains an unknown variable raised to the power of 2. The general form of a equation is:

ax^2 + bx + c = 0

In this equation:

- x represents the unknown variable.
- a, b, and c are constants, where a ≠ 0.
- a is the coefficient of x^2, b is the coefficient of x, and c is the constant term.

The equation represents a parabolic curve when graphed. It may have two real solutions, one real solution (when the discriminant is zero), or two complex solutions (when the discriminant is negative). The solutions represent the values of x that satisfy the equation and make it true. The solutions can be found using the quadratic formula:

x = (-b ± √(b^2 – 4ac)) / (2a)

The discriminant, represented by Δ, is the term under the square root in the quadratic formula:

Δ = b^2 – 4ac

The discriminant determines the nature of the solutions. If Δ > 0, there are two distinct real solutions. If Δ = 0, there is one real solution (repeated root). If Δ < 0, there are two complex solutions.

Solving quadratic equations is useful in various fields such as mathematics, physics, engineering, and computer science for modeling and analyzing various phenomena and systems.

### Equation Solver

# Quadratic Equation Solver

## Usage

Enter the coefficients of the quadratic equation (a, b, and c) to find the solutions.

### Class 10 Mathematics – Quadratic Equations

For Class 10 students, the quadratic formulas that are commonly taught are the quadratic formula for finding the roots and the formula for the vertex of a quadratic equation. Here are the formulas:

- Quadratic Formula (Roots): The quadratic formula is used to find the roots (solutions) of a quadratic equation in the form ax^2 + bx + c = 0.

The formula is:

x = (-b ± √(b^2 – 4ac)) / (2a)

Where:

- x represents the unknown variable (the roots of the quadratic equation).
- a, b, and c are the coefficients of the quadratic equation, with a ≠ 0.
- ± indicates that there are two possible solutions, one with a plus sign and one with a minus sign.

The quadratic formula can be used to find the roots of any quadratic equation, whether the discriminant is positive, zero, or negative.

- Formula for the Vertex: The vertex formula is used to find the coordinates of the vertex of a quadratic equation in the form ax^2 + bx + c.

The formula for the x-coordinate of the vertex is:

x = -b / (2a)

The formula for the y-coordinate of the vertex is:

y = f(x) = c – (b^2 / (4a))

#### Where:

- x represents the x-coordinate of the vertex.
- y represents the y-coordinate of the vertex.
- a, b, and c are the coefficients of the quadratic equation, with a ≠ 0.

The vertex represents the minimum or maximum point on the quadratic graph (parabola) depending on the coefficient of the x^2 term. If a > 0, the parabola opens upwards, and the vertex is the minimum point. If a < 0, the parabola opens downwards, and the vertex is the maximum point.

These formulas are fundamental in solving and understanding quadratic equations, and they are commonly taught in Class 10 mathematics curriculum.

### Examples :

Certainly! Here are a few examples of quadratic equations along with their solutions:

Example 1:

Solve the quadratic equation: 2x^2 – 5x + 3 = 0

Using the quadratic formula:

a = 2, b = -5, c = 3

x = (-(-5) ± √((-5)^2 – 4 * 2 * 3)) / (2 * 2)

x = (5 ± √(25 – 24)) / 4

x = (5 ± √1) / 4

So, the solutions are:

x1 = (5 + 1) / 4 = 6 / 4 = 3/2

x2 = (5 – 1) / 4 = 4 / 4 = 1

Therefore, the solutions to the quadratic equation 2x^2 – 5x + 3 = 0 are x = 3/2 and x = 1.

Example 2:

Solve the quadratic equation: x^2 + 6x + 9 = 0

Using the quadratic formula:

a = 1, b = 6, c = 9

x = (-(6) ± √((6)^2 – 4 * 1 * 9)) / (2 * 1)

x = (-6 ± √(36 – 36)) / 2

x = (-6 ± √0) / 2

Since the discriminant is zero, there is one real solution:

x = -6 / 2 = -3

Therefore, the solution to the quadratic equation x^2 + 6x + 9 = 0 is x = -3.

Example 3:

Solve the quadratic equation: 3x^2 + 4x + 2 = 0

Using the quadratic formula:

a = 3, b = 4, c = 2

x = (-(4) ± √((4)^2 – 4 * 3 * 2)) / (2 * 3)

x = (-4 ± √(16 – 24)) / 6

x = (-4 ± √(-8)) / 6

Since the discriminant is negative, the solutions are complex numbers:

x = (-4 ± 2√2i) / 6

Therefore, the solutions to the quadratic equation 3x^2 + 4x + 2 = 0 are x = (-4 + 2√2i) / 6 and x = (-4 – 2√2i) / 6.

These examples illustrate different scenarios of quadratic equations and their solutions, including cases with real solutions, repeated roots, and complex solutions.