Solve y^2-9y-18

$${ y }^{ 2 } -9y-18$$

Factor

$$\left(y-\frac{9-3\sqrt{17}}{2}\right)\left(y-\frac{3\sqrt{17}+9}{2}\right)$$

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.

$$y^{2}-9y-18=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-18\right)}}{2}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$y=\frac{-\left(-9\right)±\sqrt{81+72}}{2}$$

Add 81 to 72.

$$y=\frac{-\left(-9\right)±\sqrt{153}}{2}$$

Take the square root of 153.

$$y=\frac{-\left(-9\right)±3\sqrt{17}}{2}$$

The opposite of -9 is 9.

$$y=\frac{9±3\sqrt{17}}{2}$$

Now solve the equation y=\frac{9±3\sqrt{17}}{2} when ± is plus. Add 9 to 3\sqrt{17}.

$$y=\frac{3\sqrt{17}+9}{2}$$

Now solve the equation y=\frac{9±3\sqrt{17}}{2} when ± is minus. Subtract 3\sqrt{17} from 9.

$$y=\frac{9-3\sqrt{17}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9+3\sqrt{17}}{2} for x_{1} and \frac{9-3\sqrt{17}}{2} for x_{2}.

$$y^{2}-9y-18=\left(y-\frac{3\sqrt{17}+9}{2}\right)\left(y-\frac{9-3\sqrt{17}}{2}\right)$$

Evaluate

$$y^{2}-9y-18$$