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

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

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-1$$

Steps Using the Quadratic Formula

Add \frac{3}{2}x^{2} to both sides.

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

Add 3x to both sides.

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

Combine -x and 3x to get 2x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 2.

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

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

$$x=\frac{-2±\sqrt{4-6\times \frac{1}{2}}}{2\times \frac{3}{2}}$$

Multiply -6 times \frac{1}{2}.

$$x=\frac{-2±\sqrt{4-3}}{2\times \frac{3}{2}}$$

Add 4 to -3.

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

Take the square root of 1.

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

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

$$x=\frac{-2±1}{3}$$

Now solve the equation x=\frac{-2±1}{3} when ± is plus. Add -2 to 1.

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

Now solve the equation x=\frac{-2±1}{3} when ± is minus. Subtract 1 from -2.

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

Divide -3 by 3.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Add \frac{3}{2}x^{2} to both sides.

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

Add 3x to both sides.

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

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

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

Combine -x and 3x to get 2x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

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

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

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

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

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

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{2}{3} from both sides of the equation.

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