Solve x^2-15x-36

$${ x }^{ 2 } -15x-36$$

Factor

$$\left(x-\frac{15-3\sqrt{41}}{2}\right)\left(x-\frac{3\sqrt{41}+15}{2}\right)$$

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.

$$x^{2}-15x-36=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-36\right)}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$x=\frac{-\left(-15\right)±\sqrt{225+144}}{2}$$

Add 225 to 144.

$$x=\frac{-\left(-15\right)±\sqrt{369}}{2}$$

Take the square root of 369.

$$x=\frac{-\left(-15\right)±3\sqrt{41}}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±3\sqrt{41}}{2}$$

Now solve the equation x=\frac{15±3\sqrt{41}}{2} when ± is plus. Add 15 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}+15}{2}$$

Now solve the equation x=\frac{15±3\sqrt{41}}{2} when ± is minus. Subtract 3\sqrt{41} from 15.

$$x=\frac{15-3\sqrt{41}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{15+3\sqrt{41}}{2} for x_{1} and \frac{15-3\sqrt{41}}{2} for x_{2}.

$$x^{2}-15x-36=\left(x-\frac{3\sqrt{41}+15}{2}\right)\left(x-\frac{15-3\sqrt{41}}{2}\right)$$

Evaluate

$$x^{2}-15x-36$$