15b+9b^2=0

Solution for 15b+9b^2=0 equation:

15b+9b^2=0

a = 9; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·9·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*9}=\frac{-30}{18}
=-1+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*9}=\frac{0}{18}
=0
$