Solve 3x(x-4)=4x+(x-2)

$$3 x ( x - 4 ) = 4 x + ( x - 2 )$$

Solve for x

$$x = \frac{\sqrt{265} + 17}{6} \approx 5.546470099$$

$$x=\frac{17-\sqrt{265}}{6}\approx 0.120196567$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x-4.

$$3x^{2}-12x=4x+x-2$$

Combine 4x and x to get 5x.

$$3x^{2}-12x=5x-2$$

Subtract 5x from both sides.

$$3x^{2}-12x-5x=-2$$

Combine -12x and -5x to get -17x.

$$3x^{2}-17x=-2$$

Add 2 to both sides.

$$3x^{2}-17x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -17 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-17\right)±\sqrt{289-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-24}}{2\times 3}$$

Add 289 to -24.

$$x=\frac{-\left(-17\right)±\sqrt{265}}{2\times 3}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{265}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{17±\sqrt{265}}{6}$$

Now solve the equation x=\frac{17±\sqrt{265}}{6} when ± is plus. Add 17 to \sqrt{265}.

$$x=\frac{\sqrt{265}+17}{6}$$

Now solve the equation x=\frac{17±\sqrt{265}}{6} when ± is minus. Subtract \sqrt{265} from 17.

$$x=\frac{17-\sqrt{265}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{265}+17}{6} x=\frac{17-\sqrt{265}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x-4.

$$3x^{2}-12x=4x+x-2$$

Combine 4x and x to get 5x.

$$3x^{2}-12x=5x-2$$

Subtract 5x from both sides.

$$3x^{2}-12x-5x=-2$$

Combine -12x and -5x to get -17x.

$$3x^{2}-17x=-2$$

Divide both sides by 3.

$$\frac{3x^{2}-17x}{3}=-\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{17}{3}x=-\frac{2}{3}$$

Divide -\frac{17}{3}, the coefficient of the x term, by 2 to get -\frac{17}{6}. Then add the square of -\frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{3}x+\left(-\frac{17}{6}\right)^{2}=-\frac{2}{3}+\left(-\frac{17}{6}\right)^{2}$$

Square -\frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=-\frac{2}{3}+\frac{289}{36}$$

Add -\frac{2}{3} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=\frac{265}{36}$$

Factor x^{2}-\frac{17}{3}x+\frac{289}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{6}\right)^{2}=\frac{265}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{6}\right)^{2}}=\sqrt{\frac{265}{36}}$$

Simplify.

$$x-\frac{17}{6}=\frac{\sqrt{265}}{6} x-\frac{17}{6}=-\frac{\sqrt{265}}{6}$$

Add \frac{17}{6} to both sides of the equation.

$$x=\frac{\sqrt{265}+17}{6} x=\frac{17-\sqrt{265}}{6}$$