((3+b)*(3+b))-2b=39

Solution for ((3+b)*(3+b))-2b=39 equation:

((3+b)(3+b))-2b=39

We move all terms to the left:

((3+b)(3+b))-2b-(39)=0

We add all the numbers together, and all the variables

((b+3)(b+3))-2b-39=0

We add all the numbers together, and all the variables

-2b+((b+3)(b+3))-39=0

We multiply parentheses ..

((+b^2+3b+3b+9))-2b-39=0


We calculate terms in parentheses: +((+b^2+3b+3b+9)), so:

(+b^2+3b+3b+9)

We get rid of parentheses

b^2+3b+3b+9

We add all the numbers together, and all the variables

b^2+6b+9

Back to the equation:

+(b^2+6b+9)


We add all the numbers together, and all the variables

-2b+(b^2+6b+9)-39=0

We get rid of parentheses

b^2-2b+6b+9-39=0

We add all the numbers together, and all the variables

b^2+4b-30=0

a = 1; b = 4; c = -30;
Δ = b2-4ac
Δ = 42-4·1·(-30)
Δ = 136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{34}}{2*1}=\frac{-4-2\sqrt{34}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{34}}{2*1}=\frac{-4+2\sqrt{34}}{2}
$