Solve x^2-5x-28

$${ x }^{ 2 } -5x-28$$

Factor

$$\left(x-\frac{5-\sqrt{137}}{2}\right)\left(x-\frac{\sqrt{137}+5}{2}\right)$$

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.

$$x^{2}-5x-28=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-28\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$x=\frac{-\left(-5\right)±\sqrt{25+112}}{2}$$

Add 25 to 112.

$$x=\frac{-\left(-5\right)±\sqrt{137}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{137}}{2}$$

Now solve the equation x=\frac{5±\sqrt{137}}{2} when ± is plus. Add 5 to \sqrt{137}.

$$x=\frac{\sqrt{137}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{137}}{2} when ± is minus. Subtract \sqrt{137} from 5.

$$x=\frac{5-\sqrt{137}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{137}}{2} for x_{1} and \frac{5-\sqrt{137}}{2} for x_{2}.

$$x^{2}-5x-28=\left(x-\frac{\sqrt{137}+5}{2}\right)\left(x-\frac{5-\sqrt{137}}{2}\right)$$

Evaluate

$$x^{2}-5x-28$$