Solve 0=-1/4x^2+3/2x+4

$$0 = - \frac { 1 } { 4 } x ^ { 2 } + \frac { 3 } { 2 } x + 4$$

Solve for x

$$x=-2$$

$$x=8$$

Steps Using the Quadratic Formula

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

$$-\frac{1}{4}x^{2}+\frac{3}{2}x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, \frac{3}{2} for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\left(-\frac{1}{4}\right)\times 4}}{2\left(-\frac{1}{4}\right)}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\left(-\frac{1}{4}\right)\times 4}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{2\left(-\frac{1}{4}\right)}$$

Add \frac{9}{4} to 4.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of \frac{25}{4}.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}} when ± is plus. Add -\frac{3}{2} to \frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 1 by -\frac{1}{2} by multiplying 1 by the reciprocal of -\frac{1}{2}.

$$x=-2$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}} when ± is minus. Subtract \frac{5}{2} from -\frac{3}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -4 by -\frac{1}{2} by multiplying -4 by the reciprocal of -\frac{1}{2}.

$$x=8$$

The equation is now solved.

$$x=-2 x=8$$

Steps for Completing the Square

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

$$-\frac{1}{4}x^{2}+\frac{3}{2}x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{4}x^{2}+\frac{3}{2}x=-4$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}x^{2}+\frac{3}{2}x}{-\frac{1}{4}}=-\frac{4}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$x^{2}+\frac{\frac{3}{2}}{-\frac{1}{4}}x=-\frac{4}{-\frac{1}{4}}$$

Divide \frac{3}{2} by -\frac{1}{4} by multiplying \frac{3}{2} by the reciprocal of -\frac{1}{4}.

$$x^{2}-6x=-\frac{4}{-\frac{1}{4}}$$

Divide -4 by -\frac{1}{4} by multiplying -4 by the reciprocal of -\frac{1}{4}.

$$x^{2}-6x=16$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=16+9$$

Add 16 to 9.

$$x^{2}-6x+9=25$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-3=5 x-3=-5$$

Add 3 to both sides of the equation.

$$x=8 x=-2$$