25b^2-4=15

Solution for 25b^2-4=15 equation:

25b^2-4=15

We move all terms to the left:

25b^2-4-(15)=0

We add all the numbers together, and all the variables

25b^2-19=0

a = 25; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·25·(-19)
Δ = 1900
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:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{19}}{2*25}=\frac{0-10\sqrt{19}}{50}
=-\frac{10\sqrt{19}}{50}
=-\frac{\sqrt{19}}{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{19}}{2*25}=\frac{0+10\sqrt{19}}{50}
=\frac{10\sqrt{19}}{50}
=\frac{\sqrt{19}}{5}
$