Solve 4x^2+7x-8=x

$$4 x ^ { 2 } + 7 x - 8 = x$$

Solve for x

$$x=\frac{\sqrt{41}-3}{4}\approx 0.850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2.350781059$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$4x^{2}+7x-8-x=0$$

Combine 7x and -x to get 6x.

$$4x^{2}+6x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-6±\sqrt{36+128}}{2\times 4}$$

Add 36 to 128.

$$x=\frac{-6±\sqrt{164}}{2\times 4}$$

Take the square root of 164.

$$x=\frac{-6±2\sqrt{41}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±2\sqrt{41}}{8}$$

Now solve the equation x=\frac{-6±2\sqrt{41}}{8} when ± is plus. Add -6 to 2\sqrt{41}.

$$x=\frac{2\sqrt{41}-6}{8}$$

Divide -6+2\sqrt{41} by 8.

$$x=\frac{\sqrt{41}-3}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{41}}{8} when ± is minus. Subtract 2\sqrt{41} from -6.

$$x=\frac{-2\sqrt{41}-6}{8}$$

Divide -6-2\sqrt{41} by 8.

$$x=\frac{-\sqrt{41}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Steps for Completing the Square

Subtract x from both sides.

$$4x^{2}+7x-8-x=0$$

Combine 7x and -x to get 6x.

$$4x^{2}+6x-8=0$$

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

$$4x^{2}+6x=8$$

Divide both sides by 4.

$$\frac{4x^{2}+6x}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{6}{4}x=\frac{8}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{2}x=\frac{8}{4}$$

Divide 8 by 4.

$$x^{2}+\frac{3}{2}x=2$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=2+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=2+\frac{9}{16}$$

Add 2 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$