Solve -6x^2=3x-3

$$- 6 x ^ { 2 } = 3 x - 3$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=-1$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

$$-6x^{2}-3x=-3$$

Add 3 to both sides.

$$-6x^{2}-3x+3=0$$

Divide both sides by 3.

$$-2x^{2}-x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

Rewrite -2x^{2}-x+1 as \left(-2x^{2}+x\right)+\left(-2x+1\right).

$$\left(-2x^{2}+x\right)+\left(-2x+1\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(2x-1\right)-\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-1\right)$$

To find equation solutions, solve 2x-1=0 and -x-1=0.

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

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$-6x^{2}-3x=-3$$

Add 3 to both sides.

$$-6x^{2}-3x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -3 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)\times 3}}{2\left(-6\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 3}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 3}}{2\left(-6\right)}$$

Multiply 24 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\left(-6\right)}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\left(-6\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2\left(-6\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{3±9}{-12}$$

Now solve the equation x=\frac{3±9}{-12} when ± is plus. Add 3 to 9.

$$x=\frac{12}{-12}$$

Divide 12 by -12.

$$x=-1$$

Now solve the equation x=\frac{3±9}{-12} when ± is minus. Subtract 9 from 3.

$$x=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 3x from both sides.

$$-6x^{2}-3x=-3$$

Divide both sides by -6.

$$\frac{-6x^{2}-3x}{-6}=-\frac{3}{-6}$$

Dividing by -6 undoes the multiplication by -6.

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

Reduce the fraction \frac{-3}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{2}x=-\frac{3}{-6}$$

Reduce the fraction \frac{-3}{-6} to lowest terms by extracting and canceling out 3.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

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