Solve 3+2x-x^2=x^2-4x+3

$$3 + 2 x - x ^ { 2 } = x ^ { 2 } - 4 x + 3$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Subtract x^{2} from both sides.

$$3+2x-x^{2}-x^{2}=-4x+3$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3+2x-2x^{2}=-4x+3$$

Add 4x to both sides.

$$3+2x-2x^{2}+4x=3$$

Combine 2x and 4x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

Factor out x.

$$x\left(6-2x\right)=0$$

To find equation solutions, solve x=0 and 6-2x=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$3+2x-x^{2}-x^{2}=-4x+3$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3+2x-2x^{2}=-4x+3$$

Add 4x to both sides.

$$3+2x-2x^{2}+4x=3$$

Combine 2x and 4x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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.

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

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

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

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±6}{-4}$$

Now solve the equation x=\frac{-6±6}{-4} when ± is plus. Add -6 to 6.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-6±6}{-4} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$3+2x-x^{2}-x^{2}=-4x+3$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3+2x-2x^{2}=-4x+3$$

Add 4x to both sides.

$$3+2x-2x^{2}+4x=3$$

Combine 2x and 4x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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.

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

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=\frac{0}{-2}$$

Divide 6 by -2.

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

Divide 0 by -2.

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

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

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

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

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

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$