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

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

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=0$$

Steps Using Factoring

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

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

Factor out x.

$$x\left(4x-1\right)=0$$

To find equation solutions, solve x=0 and 4x-1=0.

$$x=0 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

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

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{1±1}{8}$$

Now solve the equation x=\frac{1±1}{8} when ± is plus. Add 1 to 1.

$$x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{1±1}{8} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{4} x=0$$

Steps for Completing the Square

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

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{1}{4}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{1}{4}x=0$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{1}{8} x-\frac{1}{8}=-\frac{1}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1}{4} x=0$$