2(3b+5)b=15

Solution for 2(3b+5)b=15 equation:

2(3b+5)b=15

We move all terms to the left:

2(3b+5)b-(15)=0

We multiply parentheses

6b^2+10b-15=0

a = 6; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·6·(-15)
Δ = 460
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:

$\sqrt{\Delta}=\sqrt{460}=\sqrt{4*115}=\sqrt{4}*\sqrt{115}=2\sqrt{115}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{115}}{2*6}=\frac{-10-2\sqrt{115}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{115}}{2*6}=\frac{-10+2\sqrt{115}}{12}
$