Solve 2x+3x^2+8x=9x^2+3x-2x

$$2x+3 { x }^{ 2 } +8x=9 { x }^{ 2 } +3x-2x$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=0$$

Steps Using Factoring

Combine 2x and 8x to get 10x.

$$10x+3x^{2}=9x^{2}+3x-2x$$

Combine 3x and -2x to get x.

$$10x+3x^{2}=9x^{2}+x$$

Subtract 9x^{2} from both sides.

$$10x+3x^{2}-9x^{2}=x$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$10x-6x^{2}=x$$

Subtract x from both sides.

$$10x-6x^{2}-x=0$$

Combine 10x and -x to get 9x.

$$9x-6x^{2}=0$$

Factor out x.

$$x\left(9-6x\right)=0$$

To find equation solutions, solve x=0 and 9-6x=0.

$$x=0 x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Combine 2x and 8x to get 10x.

$$10x+3x^{2}=9x^{2}+3x-2x$$

Combine 3x and -2x to get x.

$$10x+3x^{2}=9x^{2}+x$$

Subtract 9x^{2} from both sides.

$$10x+3x^{2}-9x^{2}=x$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$10x-6x^{2}=x$$

Subtract x from both sides.

$$10x-6x^{2}-x=0$$

Combine 10x and -x to get 9x.

$$9x-6x^{2}=0$$

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.

$$-6x^{2}+9x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}}}{2\left(-6\right)}$$

Take the square root of 9^{2}.

$$x=\frac{-9±9}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-9±9}{-12}$$

Now solve the equation x=\frac{-9±9}{-12} when ± is plus. Add -9 to 9.

$$x=\frac{0}{-12}$$

Divide 0 by -12.

$$x=0$$

Now solve the equation x=\frac{-9±9}{-12} when ± is minus. Subtract 9 from -9.

$$x=-\frac{18}{-12}$$

Reduce the fraction \frac{-18}{-12} to lowest terms by extracting and canceling out 6.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=0 x=\frac{3}{2}$$

Steps for Completing the Square

Combine 2x and 8x to get 10x.

$$10x+3x^{2}=9x^{2}+3x-2x$$

Combine 3x and -2x to get x.

$$10x+3x^{2}=9x^{2}+x$$

Subtract 9x^{2} from both sides.

$$10x+3x^{2}-9x^{2}=x$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$10x-6x^{2}=x$$

Subtract x from both sides.

$$10x-6x^{2}-x=0$$

Combine 10x and -x to get 9x.

$$9x-6x^{2}=0$$

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.

$$-6x^{2}+9x=0$$

Divide both sides by -6.

$$\frac{-6x^{2}+9x}{-6}=\frac{0}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{9}{-6}x=\frac{0}{-6}$$

Reduce the fraction \frac{9}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{3}{2}x=\frac{0}{-6}$$

Divide 0 by -6.

$$x^{2}-\frac{3}{2}x=0$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{3}{4} x-\frac{3}{4}=-\frac{3}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3}{2} x=0$$