Solve 2=-1/4x^2+5/2x

$$2 = - \frac { 1 } { 4 } x ^ { 2 } + \frac { 5 } { 2 } x$$

Solve for x

$$x=\sqrt{17}+5\approx 9.123105626$$

$$x=5-\sqrt{17}\approx 0.876894374$$

Steps Using the Quadratic Formula

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

$$-\frac{1}{4}x^{2}+\frac{5}{2}x=2$$

Subtract 2 from both sides.

$$-\frac{1}{4}x^{2}+\frac{5}{2}x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, \frac{5}{2} for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4\left(-\frac{1}{4}\right)\left(-2\right)}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-2}}{2\left(-\frac{1}{4}\right)}$$

Add \frac{25}{4} to -2.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{17}{4}}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of \frac{17}{4}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} when ± is plus. Add -\frac{5}{2} to \frac{\sqrt{17}}{2}.

$$x=\frac{\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

Divide \frac{-5+\sqrt{17}}{2} by -\frac{1}{2} by multiplying \frac{-5+\sqrt{17}}{2} by the reciprocal of -\frac{1}{2}.

$$x=5-\sqrt{17}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} when ± is minus. Subtract \frac{\sqrt{17}}{2} from -\frac{5}{2}.

$$x=\frac{-\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

Divide \frac{-5-\sqrt{17}}{2} by -\frac{1}{2} by multiplying \frac{-5-\sqrt{17}}{2} by the reciprocal of -\frac{1}{2}.

$$x=\sqrt{17}+5$$

The equation is now solved.

$$x=5-\sqrt{17} x=\sqrt{17}+5$$

Steps for Completing the Square

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

$$-\frac{1}{4}x^{2}+\frac{5}{2}x=2$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}x^{2}+\frac{5}{2}x}{-\frac{1}{4}}=\frac{2}{-\frac{1}{4}}$$

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

$$x^{2}+\frac{\frac{5}{2}}{-\frac{1}{4}}x=\frac{2}{-\frac{1}{4}}$$

Divide \frac{5}{2} by -\frac{1}{4} by multiplying \frac{5}{2} by the reciprocal of -\frac{1}{4}.

$$x^{2}-10x=\frac{2}{-\frac{1}{4}}$$

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

$$x^{2}-10x=-8$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-8+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-8+25$$

Add -8 to 25.

$$x^{2}-10x+25=17$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{17}$$

Simplify.

$$x-5=\sqrt{17} x-5=-\sqrt{17}$$

Add 5 to both sides of the equation.

$$x=\sqrt{17}+5 x=5-\sqrt{17}$$