Solve 3x-6x^2=2x^2-5x

$$3 x - 6 x ^ { 2 } = 2 x ^ { 2 } - 5 x$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

$$3x-6x^{2}-2x^{2}=-5x$$

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

$$3x-8x^{2}=-5x$$

Add 5x to both sides.

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

Combine 3x and 5x to get 8x.

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

Factor out x.

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

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

$$x=0 x=1$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$3x-6x^{2}-2x^{2}=-5x$$

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

$$3x-8x^{2}=-5x$$

Add 5x to both sides.

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

Combine 3x and 5x to get 8x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-8±8}{-16}$$

Now solve the equation x=\frac{-8±8}{-16} when ± is plus. Add -8 to 8.

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

Divide 0 by -16.

$$x=0$$

Now solve the equation x=\frac{-8±8}{-16} when ± is minus. Subtract 8 from -8.

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

Divide -16 by -16.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$3x-6x^{2}-2x^{2}=-5x$$

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

$$3x-8x^{2}=-5x$$

Add 5x to both sides.

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

Combine 3x and 5x to get 8x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

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

Divide 8 by -8.

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

Divide 0 by -8.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$