Solve (x+1)^2=2x^2+7x-5

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

Solve for x

$$x=-6$$

$$x=1$$

Steps Using Factoring By Grouping

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

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract 7x from both sides.

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

Combine 2x and -7x to get -5x.

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

Add 5 to both sides.

$$-x^{2}-5x+1+5=0$$

Add 1 and 5 to get 6.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -x^{2}-5x+6 as \left(-x^{2}+x\right)+\left(-6x+6\right).

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

Factor out x in the first and 6 in the second group.

$$x\left(-x+1\right)+6\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+6\right)$$

To find equation solutions, solve -x+1=0 and x+6=0.

$$x=1 x=-6$$

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}.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract 7x from both sides.

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

Combine 2x and -7x to get -5x.

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

Add 5 to both sides.

$$-x^{2}-5x+1+5=0$$

Add 1 and 5 to get 6.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\left(-1\right)}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±7}{-2}$$

Now solve the equation x=\frac{5±7}{-2} when ± is plus. Add 5 to 7.

$$x=\frac{12}{-2}$$

Divide 12 by -2.

$$x=-6$$

Now solve the equation x=\frac{5±7}{-2} when ± is minus. Subtract 7 from 5.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-6 x=1$$

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}.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract 7x from both sides.

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

Combine 2x and -7x to get -5x.

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

Subtract 1 from both sides.

$$-x^{2}-5x=-5-1$$

Subtract 1 from -5 to get -6.

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

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=-\frac{6}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=-\frac{6}{-1}$$

Divide -6 by -1.

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

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=1 x=-6$$