# Solve X² + 5x + 6 = 0 For X

How to solve the quadratic equation where the coeficients are a = 1, b = 5 and c = 6?

Please, to solve a 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 + 5x +6 = 0, enter a = 1, b = 5 and c = 6.

a = b = c =

In this case we have a = 1, b = 5 and c = 6. So, the equation x² + 5x + 6 = 0 has 2 real roots when solved: x₁ = -2 and x₂ = -3

## 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.

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 - 4ac

For 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:
• let's apply the quadratic formula to calculate the roots.
In this case we have a = 1, b = -5 and c = 6.
• Replacing the values
• x = (5 ± √(-5)² - 4.1.6)/2.1
• x = (5 ± √25 - 24)/2
• x₁ = (5 + 1)/2 = 6/2 = 3 and
• x₂ = (5 - 1)/2 = 4/2 = 2

### Example 2:

Lets's solve the equation x² + 2x - 3 = 0:
• In this case we have a = 1, b = 2 and c = 1.
• Replacing the values
• x = (-2 ± √2² - 4.1.1)/2.1
• x = (-2 ± √4 - 4)/2
• x = (-2 ± 0)/2
• x = (-2)/2 = -1
• x = -2/2 = -1.
• so, in this particular case, x₁ = x₂ = 1