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

$$2(x+1)(2x+3)=x(4-x)-(2x-3)2$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x+1.

$$\left(2x+2\right)\left(2x+3\right)=x\left(4-x\right)-\left(2x-3\right)\times 2$$

Use the distributive property to multiply 2x+2 by 2x+3 and combine like terms.

$$4x^{2}+10x+6=x\left(4-x\right)-\left(2x-3\right)\times 2$$

Use the distributive property to multiply x by 4-x.

$$4x^{2}+10x+6=4x-x^{2}-\left(2x-3\right)\times 2$$

Use the distributive property to multiply 2x-3 by 2.

$$4x^{2}+10x+6=4x-x^{2}-\left(4x-6\right)$$

To find the opposite of 4x-6, find the opposite of each term.

$$4x^{2}+10x+6=4x-x^{2}-4x+6$$

Combine 4x and -4x to get 0.

$$4x^{2}+10x+6=-x^{2}+6$$

Add x^{2} to both sides.

$$4x^{2}+10x+6+x^{2}=6$$

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

$$5x^{2}+10x+6=6$$

Subtract 6 from both sides.

$$5x^{2}+10x+6-6=0$$

Subtract 6 from 6 to get 0.

$$5x^{2}+10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 5}$$

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-10±10}{10}$$

Now solve the equation x=\frac{-10±10}{10} when ± is plus. Add -10 to 10.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-10±10}{10} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x+1.

$$\left(2x+2\right)\left(2x+3\right)=x\left(4-x\right)-\left(2x-3\right)\times 2$$

Use the distributive property to multiply 2x+2 by 2x+3 and combine like terms.

$$4x^{2}+10x+6=x\left(4-x\right)-\left(2x-3\right)\times 2$$

Use the distributive property to multiply x by 4-x.

$$4x^{2}+10x+6=4x-x^{2}-\left(2x-3\right)\times 2$$

Use the distributive property to multiply 2x-3 by 2.

$$4x^{2}+10x+6=4x-x^{2}-\left(4x-6\right)$$

To find the opposite of 4x-6, find the opposite of each term.

$$4x^{2}+10x+6=4x-x^{2}-4x+6$$

Combine 4x and -4x to get 0.

$$4x^{2}+10x+6=-x^{2}+6$$

Add x^{2} to both sides.

$$4x^{2}+10x+6+x^{2}=6$$

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

$$5x^{2}+10x+6=6$$

Subtract 6 from both sides.

$$5x^{2}+10x=6-6$$

Subtract 6 from 6 to get 0.

$$5x^{2}+10x=0$$

Divide both sides by 5.

$$\frac{5x^{2}+10x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{10}{5}x=\frac{0}{5}$$

Divide 10 by 5.

$$x^{2}+2x=\frac{0}{5}$$

Divide 0 by 5.

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

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

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

Square 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$