# How do i solve x²+3x- 108 = 0 [SOLUTION STEPS]

### Quadratic Equation Calculator

To solve a 2^{nd} order equation like ax² + bx + c = 0, enter or replace the coefficients a, b and c. Where, a is mandatory and nonzero. Ex.: To find the roots of the equation
x² + 3x - 108 = 0, enter
a = 1, b = 3 and c = -108.

a = b = c =

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## What is quadractic equation (In elementary algebra).

A quadratic equation is any equation having the form ax² + bx + c = 0 where x represents an unknown, and a, b, and c are just known numbers; they are called the 'numerical coefficients'. 0 is not allowed for the value of a because if a = 0, then the equation will be linear, not quadratic. The coeficient 'a' is the quadratic coefficient, 'b' the linear coefficient and 'c' the constant or free term.

## How to Solve Quadratic Equations With the Quadratic Formula (Baskara)

A way to solve quadractic equations is making use of this formula

The part (b² - 4ac) is called the "discriminant", because it can "discriminate" between the possible types of answer. If it is positive, you will get two real solutions, if it is zero you get just ONE solution, and if it is negative you get complex solutions. The 'discriminant' is represented by D or the Greek letter Delta (Δ):

Δ = b² - 4acFor the Quadratic Formula to work, you must arrange the equation in the form 'ax² + bx + c = 0' known as 'Standard Form'. Examples on how to find the coefficients:

- 1) x² + 2x - 3 = 0, a = 1, b = 2 and c = 1;
- 2) -x² + 2x + 4 = 0, a = -1, b = 2 and c = -4;
- 3) x² - x + 2-√8 = 0, a = 1, b = -1 and c = 2-√8;
- 4) x² + π = 0, a = 1, b = 0 and c = π;
- 5) x² - x = 0, a = 1, b = -1 and c = 0;

### Example 1:

Lets's solve the equation x² - 5x + 6 = 0:

a = 1, b = -5 and c = 6

Δ = b² - 4ac

Δ = (-5)² - 4.1.6 = 25 - 4.6

Δ = 25 - 24 = 1

x = -b ± √Δ2a

x = -(-5) ± √12.1

x = 5 ± √12 (general solution)

As Δ > 0, we will get two real roots, x₁ and x₂.

x₁ = 5 + √12 = 5 + 12 = 62 = 3

x₂ = 5 - √12 = 5 - 12 = 42 = 2

### Example 2:

x² + 2x + 1 = 0

a = 1, b = 2 and c = 1

Δ = b² - 4ac

Δ = 2² - 4.1.1 = 4 - 4.1

Δ = 4 - 4 = 0

x = -b ± √Δ2a

x = -2 ± √02.1

x = -2 ± √02

Δ = 0, meaning that x₁ = x₂ = x.

x = -22 = -1