Solve x^2-10x-48

$${ x }^{ 2 } -10x-48$$

Factor

$$\left(x-\left(5-\sqrt{73}\right)\right)\left(x-\left(\sqrt{73}+5\right)\right)$$

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.

$$x^{2}-10x-48=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-48\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-48\right)}}{2}$$

Multiply -4 times -48.

$$x=\frac{-\left(-10\right)±\sqrt{100+192}}{2}$$

Add 100 to 192.

$$x=\frac{-\left(-10\right)±\sqrt{292}}{2}$$

Take the square root of 292.

$$x=\frac{-\left(-10\right)±2\sqrt{73}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{73}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{73}}{2} when ± is plus. Add 10 to 2\sqrt{73}.

$$x=\frac{2\sqrt{73}+10}{2}$$

Divide 10+2\sqrt{73} by 2.

$$x=\sqrt{73}+5$$

Now solve the equation x=\frac{10±2\sqrt{73}}{2} when ± is minus. Subtract 2\sqrt{73} from 10.

$$x=\frac{10-2\sqrt{73}}{2}$$

Divide 10-2\sqrt{73} by 2.

$$x=5-\sqrt{73}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5+\sqrt{73} for x_{1} and 5-\sqrt{73} for x_{2}.

$$x^{2}-10x-48=\left(x-\left(\sqrt{73}+5\right)\right)\left(x-\left(5-\sqrt{73}\right)\right)$$

Evaluate

$$x^{2}-10x-48$$