Solve x^2+10x-16

$${ x }^{ 2 } +10x-16$$

Factor

$$\left(x-\left(-\sqrt{41}-5\right)\right)\left(x-\left(\sqrt{41}-5\right)\right)$$

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.

$$x^{2}+10x-16=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-16\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-10±\sqrt{100+64}}{2}$$

Add 100 to 64.

$$x=\frac{-10±\sqrt{164}}{2}$$

Take the square root of 164.

$$x=\frac{-10±2\sqrt{41}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{41}}{2} when ± is plus. Add -10 to 2\sqrt{41}.

$$x=\frac{2\sqrt{41}-10}{2}$$

Divide -10+2\sqrt{41} by 2.

$$x=\sqrt{41}-5$$

Now solve the equation x=\frac{-10±2\sqrt{41}}{2} when ± is minus. Subtract 2\sqrt{41} from -10.

$$x=\frac{-2\sqrt{41}-10}{2}$$

Divide -10-2\sqrt{41} by 2.

$$x=-\sqrt{41}-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5+\sqrt{41} for x_{1} and -5-\sqrt{41} for x_{2}.

$$x^{2}+10x-16=\left(x-\left(\sqrt{41}-5\right)\right)\left(x-\left(-\sqrt{41}-5\right)\right)$$

Evaluate

$$x^{2}+10x-16$$