Solve 2(s+1)^2-5(s+1)=3

$$2 ( s + 1 ) ^ { 2 } - 5 ( s + 1 ) = 3$$

Solve for s

$$s = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$s=2$$

Steps Using Factoring By Grouping

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

$$2\left(s^{2}+2s+1\right)-5\left(s+1\right)=3$$

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

$$2s^{2}+4s+2-5\left(s+1\right)=3$$

Use the distributive property to multiply -5 by s+1.

$$2s^{2}+4s+2-5s-5=3$$

Combine 4s and -5s to get -s.

$$2s^{2}-s+2-5=3$$

Subtract 5 from 2 to get -3.

$$2s^{2}-s-3=3$$

Subtract 3 from both sides.

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

Subtract 3 from -3 to get -6.

$$2s^{2}-s-6=0$$

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

$$a+b=-1 ab=2\left(-6\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite 2s^{2}-s-6 as \left(2s^{2}-4s\right)+\left(3s-6\right).

$$\left(2s^{2}-4s\right)+\left(3s-6\right)$$

Factor out 2s in the first and 3 in the second group.

$$2s\left(s-2\right)+3\left(s-2\right)$$

Factor out common term s-2 by using distributive property.

$$\left(s-2\right)\left(2s+3\right)$$

To find equation solutions, solve s-2=0 and 2s+3=0.

$$s=2 s=-\frac{3}{2}$$

Steps Using the Quadratic Formula

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

$$2\left(s^{2}+2s+1\right)-5\left(s+1\right)=3$$

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

$$2s^{2}+4s+2-5\left(s+1\right)=3$$

Use the distributive property to multiply -5 by s+1.

$$2s^{2}+4s+2-5s-5=3$$

Combine 4s and -5s to get -s.

$$2s^{2}-s+2-5=3$$

Subtract 5 from 2 to get -3.

$$2s^{2}-s-3=3$$

Subtract 3 from both sides.

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

Subtract 3 from -3 to get -6.

$$2s^{2}-s-6=0$$

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

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

Multiply -4 times 2.

$$s=\frac{-\left(-1\right)±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$s=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 2}$$

Add 1 to 48.

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

Take the square root of 49.

$$s=\frac{-\left(-1\right)±7}{2\times 2}$$

The opposite of -1 is 1.

$$s=\frac{1±7}{2\times 2}$$

Multiply 2 times 2.

$$s=\frac{1±7}{4}$$

Now solve the equation s=\frac{1±7}{4} when ± is plus. Add 1 to 7.

$$s=\frac{8}{4}$$

Divide 8 by 4.

$$s=2$$

Now solve the equation s=\frac{1±7}{4} when ± is minus. Subtract 7 from 1.

$$s=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$s=-\frac{3}{2}$$

The equation is now solved.

$$s=2 s=-\frac{3}{2}$$

Steps for Completing the Square

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

$$2\left(s^{2}+2s+1\right)-5\left(s+1\right)=3$$

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

$$2s^{2}+4s+2-5\left(s+1\right)=3$$

Use the distributive property to multiply -5 by s+1.

$$2s^{2}+4s+2-5s-5=3$$

Combine 4s and -5s to get -s.

$$2s^{2}-s+2-5=3$$

Subtract 5 from 2 to get -3.

$$2s^{2}-s-3=3$$

Add 3 to both sides.

$$2s^{2}-s=3+3$$

Add 3 and 3 to get 6.

$$2s^{2}-s=6$$

Divide both sides by 2.

$$\frac{2s^{2}-s}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$s^{2}-\frac{1}{2}s=\frac{6}{2}$$

Divide 6 by 2.

$$s^{2}-\frac{1}{2}s=3$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$s^{2}-\frac{1}{2}s+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$s^{2}-\frac{1}{2}s+\frac{1}{16}=\frac{49}{16}$$

Factor s^{2}-\frac{1}{2}s+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(s-\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$s-\frac{1}{4}=\frac{7}{4} s-\frac{1}{4}=-\frac{7}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$s=2 s=-\frac{3}{2}$$