Solve (x+1)^2-(2x+3)(x+4)=11(x-1)

$$( x + 1 ) ^ { 2 } - ( 2 x + 3 ) ( x + 4 ) = 11 ( x - 1 )$$

Solve for x

$$x=-20$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-\left(2x+3\right)\left(x+4\right)=11\left(x-1\right)$$

Use the distributive property to multiply 2x+3 by x+4 and combine like terms.

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

To find the opposite of 2x^{2}+11x+12, find the opposite of each term.

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

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

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

Combine 2x and -11x to get -9x.

$$-x^{2}-9x+1-12=11\left(x-1\right)$$

Subtract 12 from 1 to get -11.

$$-x^{2}-9x-11=11\left(x-1\right)$$

Use the distributive property to multiply 11 by x-1.

$$-x^{2}-9x-11=11x-11$$

Subtract 11x from both sides.

$$-x^{2}-9x-11-11x=-11$$

Combine -9x and -11x to get -20x.

$$-x^{2}-20x-11=-11$$

Add 11 to both sides.

$$-x^{2}-20x-11+11=0$$

Add -11 and 11 to get 0.

$$-x^{2}-20x=0$$

Factor out x.

$$x\left(-x-20\right)=0$$

To find equation solutions, solve x=0 and -x-20=0.

$$x=0 x=-20$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-\left(2x+3\right)\left(x+4\right)=11\left(x-1\right)$$

Use the distributive property to multiply 2x+3 by x+4 and combine like terms.

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

To find the opposite of 2x^{2}+11x+12, find the opposite of each term.

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

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

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

Combine 2x and -11x to get -9x.

$$-x^{2}-9x+1-12=11\left(x-1\right)$$

Subtract 12 from 1 to get -11.

$$-x^{2}-9x-11=11\left(x-1\right)$$

Use the distributive property to multiply 11 by x-1.

$$-x^{2}-9x-11=11x-11$$

Subtract 11x from both sides.

$$-x^{2}-9x-11-11x=-11$$

Combine -9x and -11x to get -20x.

$$-x^{2}-20x-11=-11$$

Add 11 to both sides.

$$-x^{2}-20x-11+11=0$$

Add -11 and 11 to get 0.

$$-x^{2}-20x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}}}{2\left(-1\right)}$$

Take the square root of \left(-20\right)^{2}.

$$x=\frac{-\left(-20\right)±20}{2\left(-1\right)}$$

The opposite of -20 is 20.

$$x=\frac{20±20}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{20±20}{-2}$$

Now solve the equation x=\frac{20±20}{-2} when ± is plus. Add 20 to 20.

$$x=\frac{40}{-2}$$

Divide 40 by -2.

$$x=-20$$

Now solve the equation x=\frac{20±20}{-2} when ± is minus. Subtract 20 from 20.

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

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-20 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-\left(2x+3\right)\left(x+4\right)=11\left(x-1\right)$$

Use the distributive property to multiply 2x+3 by x+4 and combine like terms.

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

To find the opposite of 2x^{2}+11x+12, find the opposite of each term.

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

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

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

Combine 2x and -11x to get -9x.

$$-x^{2}-9x+1-12=11\left(x-1\right)$$

Subtract 12 from 1 to get -11.

$$-x^{2}-9x-11=11\left(x-1\right)$$

Use the distributive property to multiply 11 by x-1.

$$-x^{2}-9x-11=11x-11$$

Subtract 11x from both sides.

$$-x^{2}-9x-11-11x=-11$$

Combine -9x and -11x to get -20x.

$$-x^{2}-20x-11=-11$$

Add 11 to both sides.

$$-x^{2}-20x=-11+11$$

Add -11 and 11 to get 0.

$$-x^{2}-20x=0$$

Divide both sides by -1.

$$\frac{-x^{2}-20x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{20}{-1}\right)x=\frac{0}{-1}$$

Divide -20 by -1.

$$x^{2}+20x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}+20x=0$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=10^{2}$$

Square 10.

$$x^{2}+20x+100=100$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{100}$$

Simplify.

$$x+10=10 x+10=-10$$

Subtract 10 from both sides of the equation.

$$x=0 x=-20$$