Solve 8x^2+2x-15=(x

$$8 x ^ { 2 } + 2 x - 15 = ( x$$

Solve for x

$$x = \frac{\sqrt{481} - 1}{16} \approx 1.308232012$$

$$x=\frac{-\sqrt{481}-1}{16}\approx -1.433232012$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$8x^{2}+2x-15-x=0$$

Combine 2x and -x to get x.

$$8x^{2}+x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 1 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 8\left(-15\right)}}{2\times 8}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 8\left(-15\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-1±\sqrt{1-32\left(-15\right)}}{2\times 8}$$

Multiply -32 times -15.

$$x=\frac{-1±\sqrt{1+480}}{2\times 8}$$

Add 1 to 480.

$$x=\frac{-1±\sqrt{481}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-1±\sqrt{481}}{16}$$

Now solve the equation x=\frac{-1±\sqrt{481}}{16} when ± is plus. Add -1 to \sqrt{481}.

$$x=\frac{\sqrt{481}-1}{16}$$

Now solve the equation x=\frac{-1±\sqrt{481}}{16} when ± is minus. Subtract \sqrt{481} from -1.

$$x=\frac{-\sqrt{481}-1}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{481}-1}{16} x=\frac{-\sqrt{481}-1}{16}$$

Steps for Completing the Square

Subtract x from both sides.

$$8x^{2}+2x-15-x=0$$

Combine 2x and -x to get x.

$$8x^{2}+x-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$8x^{2}+x=15$$

Divide both sides by 8.

$$\frac{8x^{2}+x}{8}=\frac{15}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{1}{8}x=\frac{15}{8}$$

Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\frac{15}{8}+\left(\frac{1}{16}\right)^{2}$$

Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{15}{8}+\frac{1}{256}$$

Add \frac{15}{8} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{481}{256}$$

Factor x^{2}+\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{481}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{481}{256}}$$

Simplify.

$$x+\frac{1}{16}=\frac{\sqrt{481}}{16} x+\frac{1}{16}=-\frac{\sqrt{481}}{16}$$

Subtract \frac{1}{16} from both sides of the equation.

$$x=\frac{\sqrt{481}-1}{16} x=\frac{-\sqrt{481}-1}{16}$$