-3b^2+15=6b^2

Solution for -3b^2+15=6b^2 equation:

-3b^2+15=6b^2

We move all terms to the left:

-3b^2+15-(6b^2)=0

We add all the numbers together, and all the variables

-9b^2+15=0

a = -9; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-9)·15
Δ = 540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-9}=\frac{0-6\sqrt{15}}{-18}
=-\frac{6\sqrt{15}}{-18}
=-\frac{\sqrt{15}}{-3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-9}=\frac{0+6\sqrt{15}}{-18}
=\frac{6\sqrt{15}}{-18}
=\frac{\sqrt{15}}{-3}
$