Solve (2+x)(-x^2+4x)-3=(3-x^2+4x)(x-1)

$$( 2 + x ) ( - x ^ { 2 } + 4 x ) - 3 = ( 3 - x ^ { 2 } + 4 x ) ( x - 1 )$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 2+x by -x^{2}+4x.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=\left(3-x^{2}+4x\right)\left(x-1\right)$$

Use the distributive property to multiply 3-x^{2}+4x by x-1 and combine like terms.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=-x-3-x^{3}+5x^{2}$$

Add x to both sides.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3+x=-3-x^{3}+5x^{2}$$

Combine 8x and x to get 9x.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3=-3-x^{3}+5x^{2}$$

Subtract -3 from both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3-\left(-3\right)=-x^{3}+5x^{2}$$

The opposite of -3 is 3.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+3=-x^{3}+5x^{2}$$

Add x^{3} to both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+3+x^{3}=5x^{2}$$

Add -3 and 3 to get 0.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}+x^{3}=5x^{2}$$

Subtract 5x^{2} from both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}+x^{3}-5x^{2}=0$$

Combine 4x^{2} and -5x^{2} to get -x^{2}.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)-x^{2}+x^{3}=0$$

Multiply 2 and -1 to get -2.

$$-2x^{2}+9x+x\left(-1\right)x^{2}-x^{2}+x^{3}=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-2x^{2}+9x+x^{3}\left(-1\right)-x^{2}+x^{3}=0$$

Combine -2x^{2} and -x^{2} to get -3x^{2}.

$$-3x^{2}+9x+x^{3}\left(-1\right)+x^{3}=0$$

Combine x^{3}\left(-1\right) and x^{3} to get 0.

$$-3x^{2}+9x=0$$

Factor out x.

$$x\left(-3x+9\right)=0$$

To find equation solutions, solve x=0 and -3x+9=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2+x by -x^{2}+4x.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=\left(3-x^{2}+4x\right)\left(x-1\right)$$

Use the distributive property to multiply 3-x^{2}+4x by x-1 and combine like terms.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=-x-3-x^{3}+5x^{2}$$

Add x to both sides.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3+x=-3-x^{3}+5x^{2}$$

Combine 8x and x to get 9x.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3=-3-x^{3}+5x^{2}$$

Subtract -3 from both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3-\left(-3\right)=-x^{3}+5x^{2}$$

The opposite of -3 is 3.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+3=-x^{3}+5x^{2}$$

Add x^{3} to both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+3+x^{3}=5x^{2}$$

Add -3 and 3 to get 0.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}+x^{3}=5x^{2}$$

Subtract 5x^{2} from both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}+x^{3}-5x^{2}=0$$

Combine 4x^{2} and -5x^{2} to get -x^{2}.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)-x^{2}+x^{3}=0$$

Multiply 2 and -1 to get -2.

$$-2x^{2}+9x+x\left(-1\right)x^{2}-x^{2}+x^{3}=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-2x^{2}+9x+x^{3}\left(-1\right)-x^{2}+x^{3}=0$$

Combine -2x^{2} and -x^{2} to get -3x^{2}.

$$-3x^{2}+9x+x^{3}\left(-1\right)+x^{3}=0$$

Combine x^{3}\left(-1\right) and x^{3} to get 0.

$$-3x^{2}+9x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 9^{2}.

$$x=\frac{-9±9}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-9±9}{-6}$$

Now solve the equation x=\frac{-9±9}{-6} when ± is plus. Add -9 to 9.

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-9±9}{-6} when ± is minus. Subtract 9 from -9.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

Use the distributive property to multiply 2+x by -x^{2}+4x.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=\left(3-x^{2}+4x\right)\left(x-1\right)$$

Use the distributive property to multiply 3-x^{2}+4x by x-1 and combine like terms.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3=-x-3-x^{3}+5x^{2}$$

Add x to both sides.

$$2\left(-x^{2}\right)+8x+x\left(-x^{2}\right)+4x^{2}-3+x=-3-x^{3}+5x^{2}$$

Combine 8x and x to get 9x.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3=-3-x^{3}+5x^{2}$$

Add x^{3} to both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+x^{3}=-3+5x^{2}$$

Subtract 5x^{2} from both sides.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)+4x^{2}-3+x^{3}-5x^{2}=-3$$

Combine 4x^{2} and -5x^{2} to get -x^{2}.

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

Add 3 to both sides.

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

Add -3 and 3 to get 0.

$$2\left(-x^{2}\right)+9x+x\left(-x^{2}\right)-x^{2}+x^{3}=0$$

Multiply 2 and -1 to get -2.

$$-2x^{2}+9x+x\left(-1\right)x^{2}-x^{2}+x^{3}=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-2x^{2}+9x+x^{3}\left(-1\right)-x^{2}+x^{3}=0$$

Combine -2x^{2} and -x^{2} to get -3x^{2}.

$$-3x^{2}+9x+x^{3}\left(-1\right)+x^{3}=0$$

Combine x^{3}\left(-1\right) and x^{3} to get 0.

$$-3x^{2}+9x=0$$

Divide both sides by -3.

$$\frac{-3x^{2}+9x}{-3}=\frac{0}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{9}{-3}x=\frac{0}{-3}$$

Divide 9 by -3.

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

Divide 0 by -3.

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

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.

$$x=3 x=0$$