Solve (w+6)^2=2w^2+17w+40

$$( w + 6 ) ^ { 2 } = 2 w ^ { 2 } + 17 w + 40$$

Solve for w

$$w=-4$$

$$w=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+6\right)^{2}.

$$w^{2}+12w+36=2w^{2}+17w+40$$

Subtract 2w^{2} from both sides.

$$w^{2}+12w+36-2w^{2}=17w+40$$

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

$$-w^{2}+12w+36=17w+40$$

Subtract 17w from both sides.

$$-w^{2}+12w+36-17w=40$$

Combine 12w and -17w to get -5w.

$$-w^{2}-5w+36=40$$

Subtract 40 from both sides.

$$-w^{2}-5w+36-40=0$$

Subtract 40 from 36 to get -4.

$$-w^{2}-5w-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -w^{2}+aw+bw-4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-1 b=-4$$

Rewrite -w^{2}-5w-4 as \left(-w^{2}-w\right)+\left(-4w-4\right).

$$\left(-w^{2}-w\right)+\left(-4w-4\right)$$

Factor out w in the first and 4 in the second group.

$$w\left(-w-1\right)+4\left(-w-1\right)$$

Factor out common term -w-1 by using distributive property.

$$\left(-w-1\right)\left(w+4\right)$$

To find equation solutions, solve -w-1=0 and w+4=0.

$$w=-1 w=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+6\right)^{2}.

$$w^{2}+12w+36=2w^{2}+17w+40$$

Subtract 2w^{2} from both sides.

$$w^{2}+12w+36-2w^{2}=17w+40$$

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

$$-w^{2}+12w+36=17w+40$$

Subtract 17w from both sides.

$$-w^{2}+12w+36-17w=40$$

Combine 12w and -17w to get -5w.

$$-w^{2}-5w+36=40$$

Subtract 40 from both sides.

$$-w^{2}-5w+36-40=0$$

Subtract 40 from 36 to get -4.

$$-w^{2}-5w-4=0$$

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

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

Square -5.

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

Multiply -4 times -1.

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

Multiply 4 times -4.

$$w=\frac{-\left(-5\right)±\sqrt{25-16}}{2\left(-1\right)}$$

Add 25 to -16.

$$w=\frac{-\left(-5\right)±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$w=\frac{-\left(-5\right)±3}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$w=\frac{5±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$w=\frac{5±3}{-2}$$

Now solve the equation w=\frac{5±3}{-2} when ± is plus. Add 5 to 3.

$$w=\frac{8}{-2}$$

Divide 8 by -2.

$$w=-4$$

Now solve the equation w=\frac{5±3}{-2} when ± is minus. Subtract 3 from 5.

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

Divide 2 by -2.

$$w=-1$$

The equation is now solved.

$$w=-4 w=-1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(w+6\right)^{2}.

$$w^{2}+12w+36=2w^{2}+17w+40$$

Subtract 2w^{2} from both sides.

$$w^{2}+12w+36-2w^{2}=17w+40$$

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

$$-w^{2}+12w+36=17w+40$$

Subtract 17w from both sides.

$$-w^{2}+12w+36-17w=40$$

Combine 12w and -17w to get -5w.

$$-w^{2}-5w+36=40$$

Subtract 36 from both sides.

$$-w^{2}-5w=40-36$$

Subtract 36 from 40 to get 4.

$$-w^{2}-5w=4$$

Divide both sides by -1.

$$\frac{-w^{2}-5w}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -5 by -1.

$$w^{2}+5w=\frac{4}{-1}$$

Divide 4 by -1.

$$w^{2}+5w=-4$$

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.

$$w^{2}+5w+\left(\frac{5}{2}\right)^{2}=-4+\left(\frac{5}{2}\right)^{2}$$

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

$$w^{2}+5w+\frac{25}{4}=-4+\frac{25}{4}$$

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

$$w^{2}+5w+\frac{25}{4}=\frac{9}{4}$$

Factor w^{2}+5w+\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}.

$$\left(w+\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$w+\frac{5}{2}=\frac{3}{2} w+\frac{5}{2}=-\frac{3}{2}$$

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

$$w=-1 w=-4$$