Solve x^2-17x-18=-5?

$$x ^ { 2 } - 17 x - 18 = - 5 ?$$

Solve for x

$$x = \frac{\sqrt{341} + 17}{2} \approx 17.733092656$$

$$x=\frac{17-\sqrt{341}}{2}\approx -0.733092656$$

Steps Using the Quadratic Formula

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.

$$x^{2}-17x-18=-5$$

Add 5 to both sides of the equation.

$$x^{2}-17x-18-\left(-5\right)=-5-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$x^{2}-17x-18-\left(-5\right)=0$$

Subtract -5 from -18.

$$x^{2}-17x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -17 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\left(-13\right)}}{2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\left(-13\right)}}{2}$$

Multiply -4 times -13.

$$x=\frac{-\left(-17\right)±\sqrt{289+52}}{2}$$

Add 289 to 52.

$$x=\frac{-\left(-17\right)±\sqrt{341}}{2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{341}}{2}$$

Now solve the equation x=\frac{17±\sqrt{341}}{2} when ± is plus. Add 17 to \sqrt{341}.

$$x=\frac{\sqrt{341}+17}{2}$$

Now solve the equation x=\frac{17±\sqrt{341}}{2} when ± is minus. Subtract \sqrt{341} from 17.

$$x=\frac{17-\sqrt{341}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{341}+17}{2} x=\frac{17-\sqrt{341}}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-17x-18=-5$$

Add 18 to both sides of the equation.

$$x^{2}-17x-18-\left(-18\right)=-5-\left(-18\right)$$

Subtracting -18 from itself leaves 0.

$$x^{2}-17x=-5-\left(-18\right)$$

Subtract -18 from -5.

$$x^{2}-17x=13$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=13+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=13+\frac{289}{4}$$

Add 13 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{341}{4}$$

Factor x^{2}-17x+\frac{289}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{341}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{341}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{\sqrt{341}}{2} x-\frac{17}{2}=-\frac{\sqrt{341}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{\sqrt{341}+17}{2} x=\frac{17-\sqrt{341}}{2}$$