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

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

Solve for x

$$x=-1$$

$$x=0$$

Steps Using Factoring

Subtract x^{2} from both sides.

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

Subtract 3x from both sides.

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

Combine 2x and -3x to get -x.

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

Factor out x.

$$x\left(-1-x\right)=0$$

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

$$x=0 x=-1$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 3x from both sides.

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

Combine 2x and -3x to get -x.

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

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.

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

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±1}{-2}$$

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

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

Divide 2 by -2.

$$x=-1$$

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

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-1 x=0$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 3x from both sides.

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

Combine 2x and -3x to get -x.

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=\frac{0}{-1}$$

Divide -1 by -1.

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

Divide 0 by -1.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=0 x=-1$$