12b^2+36b+27b^2=0

Solution for 12b^2+36b+27b^2=0 equation:

12b^2+36b+27b^2=0

We add all the numbers together, and all the variables

39b^2+36b=0

a = 39; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·39·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*39}=\frac{-72}{78}
=-12/13
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*39}=\frac{0}{78}
=0
$