12b^2+6b^2=18b

Solution for 12b^2+6b^2=18b equation:

12b^2+6b^2=18b

We move all terms to the left:

12b^2+6b^2-(18b)=0

We add all the numbers together, and all the variables

18b^2-18b=0

a = 18; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·18·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*18}=\frac{0}{36}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*18}=\frac{36}{36}
=1
$