b+(3b+10)+(3b-19)2b=360

Solution for b+(3b+10)+(3b-19)2b=360 equation:

b+(3b+10)+(3b-19)2b=360

We move all terms to the left:

b+(3b+10)+(3b-19)2b-(360)=0

We multiply parentheses

6b^2+b+(3b+10)-38b-360=0

We get rid of parentheses

6b^2+b+3b-38b+10-360=0

We add all the numbers together, and all the variables

6b^2-34b-350=0

a = 6; b = -34; c = -350;
Δ = b2-4ac
Δ = -342-4·6·(-350)
Δ = 9556
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:

$\sqrt{\Delta}=\sqrt{9556}=\sqrt{4*2389}=\sqrt{4}*\sqrt{2389}=2\sqrt{2389}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{2389}}{2*6}=\frac{34-2\sqrt{2389}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{2389}}{2*6}=\frac{34+2\sqrt{2389}}{12}
$