binomials

Solve for w:
(w + 13)^2 = (2 w + 4) (3 w + 7)

Write the quadratic polynomial on the right-hand side in standard form.

Expand out terms of the right-hand side:
(w + 13)^2 = 6 w^2 + 26 w + 28

Move everything to the left-hand side.
Subtract 6 w^2 + 26 w + 28 from both sides:
-28 - 26 w - 6 w^2 + (w + 13)^2 = 0

Write the quadratic polynomial on the left-hand side in standard form.
Expand out terms of the left-hand side:
141 - 5 w^2 = 0

Isolate terms with w to the left-hand side.
Subtract 141 from both sides:
-5 w^2 = -141

Divide both sides by a constant to simplify the equation.
Divide both sides by -5:
w^2 =± 141/5

(w + 13)²  =  (3w + 7)(2w + 4)

                                                                   First multiply out the parenthesees on both sides.

(w + 13)(w + 13)   =   (3w + 7)(2w + 4)

w² + 13w + 13w + 169   =   6w² + 12w + 14w + 28

w² + 26w + 169   =   6w² + 26w + 28

                                                                   Subtract  26w  from both sides.

w² + 169  =  6w² + 28

                                                                   Subtract  w²  from both sides.

169  =  5w² + 28

                                                                   Subtract  28  from both sides.

141  =  5w²

                                                                   Divide both sides by  5 .

141/5  =  w²

w²  =  28.2

We expand both sides to find

$\begin{align*} (w+13)(w+13)&=(3w+7)(2w+4)\\ w^2+26w+169&=3w(2w+4)+7(2w+4)\\ w^2+26w+169&=6w^2+12w+14w+28\\ w^2+26w+169&=6w^2+26w+28\\ w^2+169&=6w^2+28\\ 141&=5w^2\\ \frac{141}{5}&=w^2.\\ \end{align*}$



So, expressed as a decimal, our answer is $\frac{141}{5}=\boxed{28.2}$ .