Solve (2)2x^2+7x-4=0

$$( 2 ) 2 x ^ { 2 } + 7 x - 4 = 0$$

Solve for x

$$x=\frac{\sqrt{113}-7}{8}\approx 0.453768227$$

$$x=\frac{-\sqrt{113}-7}{8}\approx -2.203768227$$

Steps Using the Quadratic Formula

Multiply 2 and 2 to get 4.

$$4x^{2}+7x-4=0$$

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

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

Square 7.

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

Multiply -4 times 4.

$$x=\frac{-7±\sqrt{49-16\left(-4\right)}}{2\times 4}$$

Multiply -16 times -4.

$$x=\frac{-7±\sqrt{49+64}}{2\times 4}$$

Add 49 to 64.

$$x=\frac{-7±\sqrt{113}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-7±\sqrt{113}}{8}$$

Now solve the equation x=\frac{-7±\sqrt{113}}{8} when ± is plus. Add -7 to \sqrt{113}.

$$x=\frac{\sqrt{113}-7}{8}$$

Now solve the equation x=\frac{-7±\sqrt{113}}{8} when ± is minus. Subtract \sqrt{113} from -7.

$$x=\frac{-\sqrt{113}-7}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{113}-7}{8} x=\frac{-\sqrt{113}-7}{8}$$

Steps for Completing the Square

Multiply 2 and 2 to get 4.

$$4x^{2}+7x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 4 by 4.

$$x^{2}+\frac{7}{4}x=1$$

Divide \frac{7}{4}, the coefficient of the x term, by 2 to get \frac{7}{8}. Then add the square of \frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{4}x+\left(\frac{7}{8}\right)^{2}=1+\left(\frac{7}{8}\right)^{2}$$

Square \frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=1+\frac{49}{64}$$

Add 1 to \frac{49}{64}.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{113}{64}$$

Factor x^{2}+\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{8}\right)^{2}=\frac{113}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{8}\right)^{2}}=\sqrt{\frac{113}{64}}$$

Simplify.

$$x+\frac{7}{8}=\frac{\sqrt{113}}{8} x+\frac{7}{8}=-\frac{\sqrt{113}}{8}$$

Subtract \frac{7}{8} from both sides of the equation.

$$x=\frac{\sqrt{113}-7}{8} x=\frac{-\sqrt{113}-7}{8}$$