Solve 2w^2+5w+11=(w-1)^2

$$2 w ^ { 2 } + 5 w + 11 = ( w - 1 ) ^ { 2 }$$

Solve for w

$$w=-5$$

$$w=-2$$

Steps Using Factoring

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

$$2w^{2}+5w+11=w^{2}-2w+1$$

Subtract w^{2} from both sides.

$$2w^{2}+5w+11-w^{2}=-2w+1$$

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

$$w^{2}+5w+11=-2w+1$$

Add 2w to both sides.

$$w^{2}+5w+11+2w=1$$

Combine 5w and 2w to get 7w.

$$w^{2}+7w+11=1$$

Subtract 1 from both sides.

$$w^{2}+7w+11-1=0$$

Subtract 1 from 11 to get 10.

$$w^{2}+7w+10=0$$

To solve the equation, factor w^{2}+7w+10 using formula w^{2}+\left(a+b\right)w+ab=\left(w+a\right)\left(w+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=2 b=5$$

Rewrite factored expression \left(w+a\right)\left(w+b\right) using the obtained values.

$$\left(w+2\right)\left(w+5\right)$$

To find equation solutions, solve w+2=0 and w+5=0.

$$w=-2 w=-5$$

Steps Using Factoring By Grouping

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

$$2w^{2}+5w+11=w^{2}-2w+1$$

Subtract w^{2} from both sides.

$$2w^{2}+5w+11-w^{2}=-2w+1$$

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

$$w^{2}+5w+11=-2w+1$$

Add 2w to both sides.

$$w^{2}+5w+11+2w=1$$

Combine 5w and 2w to get 7w.

$$w^{2}+7w+11=1$$

Subtract 1 from both sides.

$$w^{2}+7w+11-1=0$$

Subtract 1 from 11 to get 10.

$$w^{2}+7w+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw+10. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 7.

$$a=2 b=5$$

Rewrite w^{2}+7w+10 as \left(w^{2}+2w\right)+\left(5w+10\right).

$$\left(w^{2}+2w\right)+\left(5w+10\right)$$

Factor out w in the first and 5 in the second group.

$$w\left(w+2\right)+5\left(w+2\right)$$

Factor out common term w+2 by using distributive property.

$$\left(w+2\right)\left(w+5\right)$$

To find equation solutions, solve w+2=0 and w+5=0.

$$w=-2 w=-5$$

Steps Using the Quadratic Formula

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

$$2w^{2}+5w+11=w^{2}-2w+1$$

Subtract w^{2} from both sides.

$$2w^{2}+5w+11-w^{2}=-2w+1$$

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

$$w^{2}+5w+11=-2w+1$$

Add 2w to both sides.

$$w^{2}+5w+11+2w=1$$

Combine 5w and 2w to get 7w.

$$w^{2}+7w+11=1$$

Subtract 1 from both sides.

$$w^{2}+7w+11-1=0$$

Subtract 1 from 11 to get 10.

$$w^{2}+7w+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-7±\sqrt{7^{2}-4\times 10}}{2}$$

Square 7.

$$w=\frac{-7±\sqrt{49-4\times 10}}{2}$$

Multiply -4 times 10.

$$w=\frac{-7±\sqrt{49-40}}{2}$$

Add 49 to -40.

$$w=\frac{-7±\sqrt{9}}{2}$$

Take the square root of 9.

$$w=\frac{-7±3}{2}$$

Now solve the equation w=\frac{-7±3}{2} when ± is plus. Add -7 to 3.

$$w=-\frac{4}{2}$$

Divide -4 by 2.

$$w=-2$$

Now solve the equation w=\frac{-7±3}{2} when ± is minus. Subtract 3 from -7.

$$w=-\frac{10}{2}$$

Divide -10 by 2.

$$w=-5$$

The equation is now solved.

$$w=-2 w=-5$$

Steps for Completing the Square

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

$$2w^{2}+5w+11=w^{2}-2w+1$$

Subtract w^{2} from both sides.

$$2w^{2}+5w+11-w^{2}=-2w+1$$

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

$$w^{2}+5w+11=-2w+1$$

Add 2w to both sides.

$$w^{2}+5w+11+2w=1$$

Combine 5w and 2w to get 7w.

$$w^{2}+7w+11=1$$

Subtract 11 from both sides.

$$w^{2}+7w=1-11$$

Subtract 11 from 1 to get -10.

$$w^{2}+7w=-10$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}+7w+\left(\frac{7}{2}\right)^{2}=-10+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+7w+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$w^{2}+7w+\frac{49}{4}=\frac{9}{4}$$

Factor w^{2}+7w+\frac{49}{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}.

$$\left(w+\frac{7}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$w+\frac{7}{2}=\frac{3}{2} w+\frac{7}{2}=-\frac{3}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$w=-2 w=-5$$