15b2-4b+15b=0

Solution for 15b2-4b+15b=0 equation:

15b^2-4b+15b=0

We add all the numbers together, and all the variables

15b^2+11b=0

a = 15; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·15·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*15}=\frac{-22}{30}
=-11/15
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*15}=\frac{0}{30}
=0
$