Solve 0=2x^2+7x-8

$$0 = 2 x ^ { 2 } + 7 x - 8$$

Solve for x

$$x=\frac{\sqrt{113}-7}{4}\approx 0.907536453$$

$$x=\frac{-\sqrt{113}-7}{4}\approx -4.407536453$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+7x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-7±\sqrt{49+64}}{2\times 2}$$

Add 49 to 64.

$$x=\frac{-7±\sqrt{113}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±\sqrt{113}}{4}$$

Now solve the equation x=\frac{-7±\sqrt{113}}{4} when ± is plus. Add -7 to \sqrt{113}.

$$x=\frac{\sqrt{113}-7}{4}$$

Now solve the equation x=\frac{-7±\sqrt{113}}{4} when ± is minus. Subtract \sqrt{113} from -7.

$$x=\frac{-\sqrt{113}-7}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{113}-7}{4} x=\frac{-\sqrt{113}-7}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+7x-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$2x^{2}+7x=8$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=\frac{8}{2}$$

Divide 8 by 2.

$$x^{2}+\frac{7}{2}x=4$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=4+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=4+\frac{49}{16}$$

Add 4 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{113}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{113}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{113}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{\sqrt{113}}{4} x+\frac{7}{4}=-\frac{\sqrt{113}}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{\sqrt{113}-7}{4} x=\frac{-\sqrt{113}-7}{4}$$