Solve 3x^2-8x+4x/x-2=5x+8

$$\frac{ 3 { x }^{ 2 } -8x+4x }{ x-2 } = 5x+8$$

Solve for x

$$x = \frac{\sqrt{33} - 1}{2} \approx 2.372281323$$

$$x=\frac{-\sqrt{33}-1}{2}\approx -3.372281323$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$3x^{2}-8x+4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Combine -8x and 4x to get -4x.

$$3x^{2}-4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Use the distributive property to multiply 5x by x-2.

$$3x^{2}-4x=5x^{2}-10x+\left(x-2\right)\times 8$$

Use the distributive property to multiply x-2 by 8.

$$3x^{2}-4x=5x^{2}-10x+8x-16$$

Combine -10x and 8x to get -2x.

$$3x^{2}-4x=5x^{2}-2x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-4x-5x^{2}=-2x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-4x=-2x-16$$

Add 2x to both sides.

$$-2x^{2}-4x+2x=-16$$

Combine -4x and 2x to get -2x.

$$-2x^{2}-2x=-16$$

Add 16 to both sides.

$$-2x^{2}-2x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -2 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times -2.

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

Multiply 8 times 16.

$$x=\frac{-\left(-2\right)±\sqrt{4+128}}{2\left(-2\right)}$$

Add 4 to 128.

$$x=\frac{-\left(-2\right)±\sqrt{132}}{2\left(-2\right)}$$

Take the square root of 132.

$$x=\frac{-\left(-2\right)±2\sqrt{33}}{2\left(-2\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{33}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{33}}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{33}}{-4} when ± is plus. Add 2 to 2\sqrt{33}.

$$x=\frac{2\sqrt{33}+2}{-4}$$

Divide 2+2\sqrt{33} by -4.

$$x=\frac{-\sqrt{33}-1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{33}}{-4} when ± is minus. Subtract 2\sqrt{33} from 2.

$$x=\frac{2-2\sqrt{33}}{-4}$$

Divide 2-2\sqrt{33} by -4.

$$x=\frac{\sqrt{33}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{33}-1}{2} x=\frac{\sqrt{33}-1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$3x^{2}-8x+4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Combine -8x and 4x to get -4x.

$$3x^{2}-4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Use the distributive property to multiply 5x by x-2.

$$3x^{2}-4x=5x^{2}-10x+\left(x-2\right)\times 8$$

Use the distributive property to multiply x-2 by 8.

$$3x^{2}-4x=5x^{2}-10x+8x-16$$

Combine -10x and 8x to get -2x.

$$3x^{2}-4x=5x^{2}-2x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-4x-5x^{2}=-2x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-4x=-2x-16$$

Add 2x to both sides.

$$-2x^{2}-4x+2x=-16$$

Combine -4x and 2x to get -2x.

$$-2x^{2}-2x=-16$$

Divide both sides by -2.

$$\frac{-2x^{2}-2x}{-2}=-\frac{16}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{2}{-2}\right)x=-\frac{16}{-2}$$

Divide -2 by -2.

$$x^{2}+x=-\frac{16}{-2}$$

Divide -16 by -2.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=8+\frac{1}{4}$$

Add 8 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{33}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{33}}{2} x+\frac{1}{2}=-\frac{\sqrt{33}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{33}-1}{2} x=\frac{-\sqrt{33}-1}{2}$$