Solve -3=3/8x^2-3/4x-3

$$- 3 = \frac { 3 } { 8 } x ^ { 2 } - \frac { 3 } { 4 } x - 3$$

Solve for x

$$x=2$$

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{8}x^{2}-\frac{3}{4}x-3=-3$$

Add 3 to both sides.

$$\frac{3}{8}x^{2}-\frac{3}{4}x-3+3=0$$

Add -3 and 3 to get 0.

$$\frac{3}{8}x^{2}-\frac{3}{4}x=0$$

Factor out x.

$$x\left(\frac{3}{8}x-\frac{3}{4}\right)=0$$

To find equation solutions, solve x=0 and \frac{3x}{8}-\frac{3}{4}=0.

$$x=0 x=2$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{8}x^{2}-\frac{3}{4}x-3=-3$$

Add 3 to both sides.

$$\frac{3}{8}x^{2}-\frac{3}{4}x-3+3=0$$

Add -3 and 3 to get 0.

$$\frac{3}{8}x^{2}-\frac{3}{4}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{8} for a, -\frac{3}{4} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{4}\right)±\sqrt{\left(-\frac{3}{4}\right)^{2}}}{2\times \frac{3}{8}}$$

Take the square root of \left(-\frac{3}{4}\right)^{2}.

$$x=\frac{-\left(-\frac{3}{4}\right)±\frac{3}{4}}{2\times \frac{3}{8}}$$

The opposite of -\frac{3}{4} is \frac{3}{4}.

$$x=\frac{\frac{3}{4}±\frac{3}{4}}{2\times \frac{3}{8}}$$

Multiply 2 times \frac{3}{8}.

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

Now solve the equation x=\frac{\frac{3}{4}±\frac{3}{4}}{\frac{3}{4}} when ± is plus. Add \frac{3}{4} to \frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{3}{2}}{\frac{3}{4}}$$

Divide \frac{3}{2} by \frac{3}{4} by multiplying \frac{3}{2} by the reciprocal of \frac{3}{4}.

$$x=2$$

Now solve the equation x=\frac{\frac{3}{4}±\frac{3}{4}}{\frac{3}{4}} when ± is minus. Subtract \frac{3}{4} from \frac{3}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{3}{4}}$$

Divide 0 by \frac{3}{4} by multiplying 0 by the reciprocal of \frac{3}{4}.

$$x=0$$

The equation is now solved.

$$x=2 x=0$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{8}x^{2}-\frac{3}{4}x-3=-3$$

Add 3 to both sides.

$$\frac{3}{8}x^{2}-\frac{3}{4}x=-3+3$$

Add -3 and 3 to get 0.

$$\frac{3}{8}x^{2}-\frac{3}{4}x=0$$

Divide both sides of the equation by \frac{3}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{8}x^{2}-\frac{3}{4}x}{\frac{3}{8}}=\frac{0}{\frac{3}{8}}$$

Dividing by \frac{3}{8} undoes the multiplication by \frac{3}{8}.

$$x^{2}+\left(-\frac{\frac{3}{4}}{\frac{3}{8}}\right)x=\frac{0}{\frac{3}{8}}$$

Divide -\frac{3}{4} by \frac{3}{8} by multiplying -\frac{3}{4} by the reciprocal of \frac{3}{8}.

$$x^{2}-2x=\frac{0}{\frac{3}{8}}$$

Divide 0 by \frac{3}{8} by multiplying 0 by the reciprocal of \frac{3}{8}.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=1$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=1 x-1=-1$$

Add 1 to both sides of the equation.

$$x=2 x=0$$