Solve 4x^2+3x=2x^2

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

$$4x^{2}+3x-2x^{2}=0$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

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

Factor out x.

$$x\left(2x+3\right)=0$$

To find equation solutions, solve x=0 and 2x+3=0.

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

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$4x^{2}+3x-2x^{2}=0$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

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

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

$$x=\frac{-3±\sqrt{3^{2}}}{2\times 2}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±3}{4}$$

Now solve the equation x=\frac{-3±3}{4} when ± is plus. Add -3 to 3.

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

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-3±3}{4} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$4x^{2}+3x-2x^{2}=0$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 0 by 2.

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

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.

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

Subtract \frac{3}{4} from both sides of the equation.

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