Solve x^2-5+12x

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

Factor

$$\left(x-\left(-\sqrt{41}-6\right)\right)\left(x-\left(\sqrt{41}-6\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-5\right)}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-12±\sqrt{144+20}}{2}$$

Add 144 to 20.

$$x=\frac{-12±\sqrt{164}}{2}$$

Take the square root of 164.

$$x=\frac{-12±2\sqrt{41}}{2}$$

Now solve the equation x=\frac{-12±2\sqrt{41}}{2} when ± is plus. Add -12 to 2\sqrt{41}.

$$x=\frac{2\sqrt{41}-12}{2}$$

Divide -12+2\sqrt{41} by 2.

$$x=\sqrt{41}-6$$

Now solve the equation x=\frac{-12±2\sqrt{41}}{2} when ± is minus. Subtract 2\sqrt{41} from -12.

$$x=\frac{-2\sqrt{41}-12}{2}$$

Divide -12-2\sqrt{41} by 2.

$$x=-\sqrt{41}-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6+\sqrt{41} for x_{1} and -6-\sqrt{41} for x_{2}.

$$x^{2}+12x-5=\left(x-\left(\sqrt{41}-6\right)\right)\left(x-\left(-\sqrt{41}-6\right)\right)$$

Evaluate

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