Solve 5x^2-8x+3(x+2)x=0

$$5 x ^ { 2 } - 8 x + 3 ( x + 2 ) x = 0$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 3 by x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Use the distributive property to multiply 3x+6 by x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Combine 5x^{2} and 3x^{2} to get 8x^{2}.

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

Combine -8x and 6x to get -2x.

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

Factor out x.

$$x\left(8x-2\right)=0$$

To find equation solutions, solve x=0 and 8x-2=0.

$$x=0 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Use the distributive property to multiply 3x+6 by x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Combine 5x^{2} and 3x^{2} to get 8x^{2}.

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

Combine -8x and 6x to get -2x.

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

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

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$x=\frac{2±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{2±2}{16}$$

Now solve the equation x=\frac{2±2}{16} when ± is plus. Add 2 to 2.

$$x=\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{2±2}{16} when ± is minus. Subtract 2 from 2.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{4} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 3 by x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Use the distributive property to multiply 3x+6 by x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Combine 5x^{2} and 3x^{2} to get 8x^{2}.

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

Combine -8x and 6x to get -2x.

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

Divide both sides by 8.

$$\frac{8x^{2}-2x}{8}=\frac{0}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{2}{8}\right)x=\frac{0}{8}$$

Reduce the fraction \frac{-2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{4}x=\frac{0}{8}$$

Divide 0 by 8.

$$x^{2}-\frac{1}{4}x=0$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-\frac{1}{4}x+\frac{1}{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}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{1}{8} x-\frac{1}{8}=-\frac{1}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1}{4} x=0$$