Solve -3x^2-3=3x-5

$$-3 { x }^{ 2 } -3=3x-5$$

Solve for x

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}\approx 0.457427108$$

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}\approx -1.457427108$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$-3x^{2}-3-3x=-5$$

Add 5 to both sides.

$$-3x^{2}-3-3x+5=0$$

Add -3 and 5 to get 2.

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

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.

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

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

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

Square -3.

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

Multiply -4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

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

Add 9 to 24.

$$x=\frac{-\left(-3\right)±\sqrt{33}}{2\left(-3\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{33}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{3±\sqrt{33}}{-6}$$

Now solve the equation x=\frac{3±\sqrt{33}}{-6} when ± is plus. Add 3 to \sqrt{33}.

$$x=\frac{\sqrt{33}+3}{-6}$$

Divide 3+\sqrt{33} by -6.

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{33}}{-6} when ± is minus. Subtract \sqrt{33} from 3.

$$x=\frac{3-\sqrt{33}}{-6}$$

Divide 3-\sqrt{33} by -6.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2} x=\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$-3x^{2}-3-3x=-5$$

Add 3 to both sides.

$$-3x^{2}-3x=-5+3$$

Add -5 and 3 to get -2.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -3 by -3.

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

Divide -2 by -3.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{2}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{11}{12}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{11}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{11}{12}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{33}}{6} x+\frac{1}{2}=-\frac{\sqrt{33}}{6}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2} x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$