Solve 20x^2+5x=4x

$$20 x ^ { 2 } + 5 x = 4 x$$

Solve for x

$$x=-\frac{1}{20}=-0.05$$

$$x=0$$

Steps Using Factoring

Subtract 4x from both sides.

$$20x^{2}+5x-4x=0$$

Combine 5x and -4x to get x.

$$20x^{2}+x=0$$

Factor out x.

$$x\left(20x+1\right)=0$$

To find equation solutions, solve x=0 and 20x+1=0.

$$x=0 x=-\frac{1}{20}$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$20x^{2}+5x-4x=0$$

Combine 5x and -4x to get x.

$$20x^{2}+x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 20}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-1±1}{40}$$

Now solve the equation x=\frac{-1±1}{40} when ± is plus. Add -1 to 1.

$$x=\frac{0}{40}$$

Divide 0 by 40.

$$x=0$$

Now solve the equation x=\frac{-1±1}{40} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{40}$$

Reduce the fraction \frac{-2}{40} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{20}$$

The equation is now solved.

$$x=0 x=-\frac{1}{20}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$20x^{2}+5x-4x=0$$

Combine 5x and -4x to get x.

$$20x^{2}+x=0$$

Divide both sides by 20.

$$\frac{20x^{2}+x}{20}=\frac{0}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{1}{20}x=\frac{0}{20}$$

Divide 0 by 20.

$$x^{2}+\frac{1}{20}x=0$$

Divide \frac{1}{20}, the coefficient of the x term, by 2 to get \frac{1}{40}. Then add the square of \frac{1}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{20}x+\left(\frac{1}{40}\right)^{2}=\left(\frac{1}{40}\right)^{2}$$

Square \frac{1}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{20}x+\frac{1}{1600}=\frac{1}{1600}$$

Factor x^{2}+\frac{1}{20}x+\frac{1}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{40}\right)^{2}=\frac{1}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{40}\right)^{2}}=\sqrt{\frac{1}{1600}}$$

Simplify.

$$x+\frac{1}{40}=\frac{1}{40} x+\frac{1}{40}=-\frac{1}{40}$$

Subtract \frac{1}{40} from both sides of the equation.

$$x=0 x=-\frac{1}{20}$$