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

$$\left. \begin{array}{l}{ ( x + 1 ) = 2 ( x + 1 ) ^ { 2 } + 3 ( x + 1 ) - 4 }\\{ = 2 ( x ^ { 2 } + 1 x + 11 + 3 x + 3 - 4 }\end{array} \right.$$

Solve for x

$$x=-3$$

$$x=0$$

Steps Using Factoring

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

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

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

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

Use the distributive property to multiply 3 by x+1.

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

Combine 4x and 3x to get 7x.

$$x+1=2x^{2}+7x+2+3-4$$

Add 2 and 3 to get 5.

$$x+1=2x^{2}+7x+5-4$$

Subtract 4 from 5 to get 1.

$$x+1=2x^{2}+7x+1$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=7x+1$$

Subtract 7x from both sides.

$$x+1-2x^{2}-7x=1$$

Combine x and -7x to get -6x.

$$-6x+1-2x^{2}=1$$

Subtract 1 from both sides.

$$-6x+1-2x^{2}-1=0$$

Subtract 1 from 1 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

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

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

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

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

Use the distributive property to multiply 3 by x+1.

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

Combine 4x and 3x to get 7x.

$$x+1=2x^{2}+7x+2+3-4$$

Add 2 and 3 to get 5.

$$x+1=2x^{2}+7x+5-4$$

Subtract 4 from 5 to get 1.

$$x+1=2x^{2}+7x+1$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=7x+1$$

Subtract 7x from both sides.

$$x+1-2x^{2}-7x=1$$

Combine x and -7x to get -6x.

$$-6x+1-2x^{2}=1$$

Subtract 1 from both sides.

$$-6x+1-2x^{2}-1=0$$

Subtract 1 from 1 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}.

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

Take the square root of \left(-6\right)^{2}.

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

The opposite of -6 is 6.

$$x=\frac{6±6}{2\left(-2\right)}$$

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.

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

Divide 12 by -4.

$$x=-3$$

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

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

Divide 0 by -4.

$$x=0$$

The equation is now solved.

$$x=-3 x=0$$

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}.

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

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

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

Use the distributive property to multiply 3 by x+1.

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

Combine 4x and 3x to get 7x.

$$x+1=2x^{2}+7x+2+3-4$$

Add 2 and 3 to get 5.

$$x+1=2x^{2}+7x+5-4$$

Subtract 4 from 5 to get 1.

$$x+1=2x^{2}+7x+1$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=7x+1$$

Subtract 7x from both sides.

$$x+1-2x^{2}-7x=1$$

Combine x and -7x to get -6x.

$$-6x+1-2x^{2}=1$$

Subtract 1 from both sides.

$$-6x-2x^{2}=1-1$$

Subtract 1 from 1 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.

$$x^{2}+\left(-\frac{6}{-2}\right)x=\frac{0}{-2}$$

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.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=0 x=-3$$