Solve 4=(-4x^2+1)+(-x+3)

$$4 = ( - 4 x ^ { 2 } + 1 ) + ( - x + 3 )$$

Solve for x

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

$$x=0$$

Steps Using the Quadratic Formula

Add 1 and 3 to get 4.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 4 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}.

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-4\right)}$$

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

Add 1 and 3 to get 4.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from 4 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.

$$x^{2}+\left(-\frac{1}{-4}\right)x=\frac{0}{-4}$$

Divide -1 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.

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

Subtract \frac{1}{8} from both sides of the equation.

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