Solve 3x(2x-5)-4x(5-2x)=0

$$3 x ( 2 x - 5 ) - 4 x ( 5 - 2 x ) = 0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 3x by 2x-5.

$$6x^{2}-15x-4x\left(5-2x\right)=0$$

Use the distributive property to multiply -4x by 5-2x.

$$6x^{2}-15x-20x+8x^{2}=0$$

Combine -15x and -20x to get -35x.

$$6x^{2}-35x+8x^{2}=0$$

Combine 6x^{2} and 8x^{2} to get 14x^{2}.

$$14x^{2}-35x=0$$

Factor out x.

$$x\left(14x-35\right)=0$$

To find equation solutions, solve x=0 and 14x-35=0.

$$x=0 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 2x-5.

$$6x^{2}-15x-4x\left(5-2x\right)=0$$

Use the distributive property to multiply -4x by 5-2x.

$$6x^{2}-15x-20x+8x^{2}=0$$

Combine -15x and -20x to get -35x.

$$6x^{2}-35x+8x^{2}=0$$

Combine 6x^{2} and 8x^{2} to get 14x^{2}.

$$14x^{2}-35x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -35 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}}}{2\times 14}$$

Take the square root of \left(-35\right)^{2}.

$$x=\frac{-\left(-35\right)±35}{2\times 14}$$

The opposite of -35 is 35.

$$x=\frac{35±35}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{35±35}{28}$$

Now solve the equation x=\frac{35±35}{28} when ± is plus. Add 35 to 35.

$$x=\frac{70}{28}$$

Reduce the fraction \frac{70}{28} to lowest terms by extracting and canceling out 14.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{35±35}{28} when ± is minus. Subtract 35 from 35.

$$x=\frac{0}{28}$$

Divide 0 by 28.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{2} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 2x-5.

$$6x^{2}-15x-4x\left(5-2x\right)=0$$

Use the distributive property to multiply -4x by 5-2x.

$$6x^{2}-15x-20x+8x^{2}=0$$

Combine -15x and -20x to get -35x.

$$6x^{2}-35x+8x^{2}=0$$

Combine 6x^{2} and 8x^{2} to get 14x^{2}.

$$14x^{2}-35x=0$$

Divide both sides by 14.

$$\frac{14x^{2}-35x}{14}=\frac{0}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$x^{2}+\left(-\frac{35}{14}\right)x=\frac{0}{14}$$

Reduce the fraction \frac{-35}{14} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{5}{2}x=\frac{0}{14}$$

Divide 0 by 14.

$$x^{2}-\frac{5}{2}x=0$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5}{2} x=0$$