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

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

Solve for w

$$w=-6$$

$$w=5$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+5\right)^{2}.

$$2w^{2}+11w-5=w^{2}+10w+25$$

Subtract w^{2} from both sides.

$$2w^{2}+11w-5-w^{2}=10w+25$$

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

$$w^{2}+11w-5=10w+25$$

Subtract 10w from both sides.

$$w^{2}+11w-5-10w=25$$

Combine 11w and -10w to get w.

$$w^{2}+w-5=25$$

Subtract 25 from both sides.

$$w^{2}+w-5-25=0$$

Subtract 25 from -5 to get -30.

$$w^{2}+w-30=0$$

To solve the equation, factor w^{2}+w-30 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.

$$a+b=1 ab=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

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

$$\left(w-5\right)\left(w+6\right)$$

To find equation solutions, solve w-5=0 and w+6=0.

$$w=5 w=-6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+5\right)^{2}.

$$2w^{2}+11w-5=w^{2}+10w+25$$

Subtract w^{2} from both sides.

$$2w^{2}+11w-5-w^{2}=10w+25$$

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

$$w^{2}+11w-5=10w+25$$

Subtract 10w from both sides.

$$w^{2}+11w-5-10w=25$$

Combine 11w and -10w to get w.

$$w^{2}+w-5=25$$

Subtract 25 from both sides.

$$w^{2}+w-5-25=0$$

Subtract 25 from -5 to get -30.

$$w^{2}+w-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite w^{2}+w-30 as \left(w^{2}-5w\right)+\left(6w-30\right).

$$\left(w^{2}-5w\right)+\left(6w-30\right)$$

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

$$w\left(w-5\right)+6\left(w-5\right)$$

Factor out common term w-5 by using distributive property.

$$\left(w-5\right)\left(w+6\right)$$

To find equation solutions, solve w-5=0 and w+6=0.

$$w=5 w=-6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+5\right)^{2}.

$$2w^{2}+11w-5=w^{2}+10w+25$$

Subtract w^{2} from both sides.

$$2w^{2}+11w-5-w^{2}=10w+25$$

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

$$w^{2}+11w-5=10w+25$$

Subtract 10w from both sides.

$$w^{2}+11w-5-10w=25$$

Combine 11w and -10w to get w.

$$w^{2}+w-5=25$$

Subtract 25 from both sides.

$$w^{2}+w-5-25=0$$

Subtract 25 from -5 to get -30.

$$w^{2}+w-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-1±\sqrt{1^{2}-4\left(-30\right)}}{2}$$

Square 1.

$$w=\frac{-1±\sqrt{1-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$w=\frac{-1±\sqrt{1+120}}{2}$$

Add 1 to 120.

$$w=\frac{-1±\sqrt{121}}{2}$$

Take the square root of 121.

$$w=\frac{-1±11}{2}$$

Now solve the equation w=\frac{-1±11}{2} when ± is plus. Add -1 to 11.

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

Divide 10 by 2.

$$w=5$$

Now solve the equation w=\frac{-1±11}{2} when ± is minus. Subtract 11 from -1.

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

Divide -12 by 2.

$$w=-6$$

The equation is now solved.

$$w=5 w=-6$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+5\right)^{2}.

$$2w^{2}+11w-5=w^{2}+10w+25$$

Subtract w^{2} from both sides.

$$2w^{2}+11w-5-w^{2}=10w+25$$

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

$$w^{2}+11w-5=10w+25$$

Subtract 10w from both sides.

$$w^{2}+11w-5-10w=25$$

Combine 11w and -10w to get w.

$$w^{2}+w-5=25$$

Add 5 to both sides.

$$w^{2}+w=25+5$$

Add 25 and 5 to get 30.

$$w^{2}+w=30$$

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

$$w^{2}+w+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+w+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$w^{2}+w+\frac{1}{4}=\frac{121}{4}$$

Factor w^{2}+w+\frac{1}{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}.

$$\left(w+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$w+\frac{1}{2}=\frac{11}{2} w+\frac{1}{2}=-\frac{11}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$w=5 w=-6$$