-5b(3b=12)=-3b-5(b+12)

Solution for -5b(3b=12)=-3b-5(b+12) equation:

-5b(3b=12)=-3b-5(b+12)

We move all terms to the left:

-5b(3b-(12))=0

We multiply parentheses

-15b^2+60b=0

a = -15; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·(-15)·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*-15}=\frac{-120}{-30}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*-15}=\frac{0}{-30}
=0
$