Solve -x^2+x+2=1/2*(-1/2x^2+3/2x+2)

$$- { x }^{ 2 } +x+2= \frac{ 1 }{ 2 } \times (- \frac{ 1 }{ 2 } { x }^{ 2 } + \frac{ 3 }{ 2 } x+2)$$

Solve for x

$$x=-1$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2} by -\frac{1}{2}x^{2}+\frac{3}{2}x+2.

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

Add \frac{1}{4}x^{2} to both sides.

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

Subtract \frac{3}{4}x from both sides.

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

Combine x and -\frac{3}{4}x to get \frac{1}{4}x.

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

Subtract 1 from both sides.

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

Subtract 1 from 2 to get 1.

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

Combine -x^{2} and \frac{1}{4}x^{2} to get -\frac{3}{4}x^{2}.

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

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

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1}{4}±\sqrt{\frac{1}{16}-4\left(-\frac{3}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply -4 times -\frac{3}{4}.

$$x=\frac{-\frac{1}{4}±\sqrt{\frac{1}{16}+3}}{2\left(-\frac{3}{4}\right)}$$

Add \frac{1}{16} to 3.

$$x=\frac{-\frac{1}{4}±\sqrt{\frac{49}{16}}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of \frac{49}{16}.

$$x=\frac{-\frac{1}{4}±\frac{7}{4}}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$x=\frac{-\frac{1}{4}±\frac{7}{4}}{-\frac{3}{2}}$$

Now solve the equation x=\frac{-\frac{1}{4}±\frac{7}{4}}{-\frac{3}{2}} when ± is plus. Add -\frac{1}{4} to \frac{7}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=-1$$

Now solve the equation x=\frac{-\frac{1}{4}±\frac{7}{4}}{-\frac{3}{2}} when ± is minus. Subtract \frac{7}{4} from -\frac{1}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

The equation is now solved.

$$x=-1 x=\frac{4}{3}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2} by -\frac{1}{2}x^{2}+\frac{3}{2}x+2.

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

Add \frac{1}{4}x^{2} to both sides.

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

Subtract \frac{3}{4}x from both sides.

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

Combine x and -\frac{3}{4}x to get \frac{1}{4}x.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

Combine -x^{2} and \frac{1}{4}x^{2} to get -\frac{3}{4}x^{2}.

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

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

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

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

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

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

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{7}{6} x-\frac{1}{6}=-\frac{7}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{4}{3} x=-1$$