Solve x^2+12x+36=5

$$x ^ { 2 } + 12 x + 36 = 5$$

Solve for x (complex solution)

$$x=\sqrt{5}-6\approx -3.763932023$$

$$x=-\left(\sqrt{5}+6\right)\approx -8.236067977$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+12x+36=5$$

Subtract 5 from both sides of the equation.

$$x^{2}+12x+36-5=5-5$$

Subtracting 5 from itself leaves 0.

$$x^{2}+12x+36-5=0$$

Subtract 5 from 36.

$$x^{2}+12x+31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 31}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 31}}{2}$$

Multiply -4 times 31.

$$x=\frac{-12±\sqrt{144-124}}{2}$$

Add 144 to -124.

$$x=\frac{-12±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-12±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{2} when ± is plus. Add -12 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-12}{2}$$

Divide -12+2\sqrt{5} by 2.

$$x=\sqrt{5}-6$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -12.

$$x=\frac{-2\sqrt{5}-12}{2}$$

Divide -12-2\sqrt{5} by 2.

$$x=-\sqrt{5}-6$$

The equation is now solved.

$$x=\sqrt{5}-6 x=-\sqrt{5}-6$$

Steps for Completing the Square

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+6=\sqrt{5} x+6=-\sqrt{5}$$

Subtract 6 from both sides of the equation.

$$x=\sqrt{5}-6 x=-\sqrt{5}-6$$

Solve for x

$$x=\sqrt{5}-6\approx -3.763932023$$

$$x=-\sqrt{5}-6\approx -8.236067977$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+12x+36=5$$

Subtract 5 from both sides of the equation.

$$x^{2}+12x+36-5=5-5$$

Subtracting 5 from itself leaves 0.

$$x^{2}+12x+36-5=0$$

Subtract 5 from 36.

$$x^{2}+12x+31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 31}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 31}}{2}$$

Multiply -4 times 31.

$$x=\frac{-12±\sqrt{144-124}}{2}$$

Add 144 to -124.

$$x=\frac{-12±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-12±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{2} when ± is plus. Add -12 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-12}{2}$$

Divide -12+2\sqrt{5} by 2.

$$x=\sqrt{5}-6$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -12.

$$x=\frac{-2\sqrt{5}-12}{2}$$

Divide -12-2\sqrt{5} by 2.

$$x=-\sqrt{5}-6$$

The equation is now solved.

$$x=\sqrt{5}-6 x=-\sqrt{5}-6$$

Steps for Completing the Square

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+6=\sqrt{5} x+6=-\sqrt{5}$$

Subtract 6 from both sides of the equation.

$$x=\sqrt{5}-6 x=-\sqrt{5}-6$$