b+(3b+10)+(3b-10)2b=360

Solution for b+(3b+10)+(3b-10)2b=360 equation:

b+(3b+10)+(3b-10)2b=360

We move all terms to the left:

b+(3b+10)+(3b-10)2b-(360)=0

We multiply parentheses

6b^2+b+(3b+10)-20b-360=0

We get rid of parentheses

6b^2+b+3b-20b+10-360=0

We add all the numbers together, and all the variables

6b^2-16b-350=0

a = 6; b = -16; c = -350;
Δ = b2-4ac
Δ = -162-4·6·(-350)
Δ = 8656
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:

$\sqrt{\Delta}=\sqrt{8656}=\sqrt{16*541}=\sqrt{16}*\sqrt{541}=4\sqrt{541}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{541}}{2*6}=\frac{16-4\sqrt{541}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{541}}{2*6}=\frac{16+4\sqrt{541}}{12}
$