48=b^2+25b^2

Solution for 48=b^2+25b^2 equation:

48=b^2+25b^2

We move all terms to the left:

48-(b^2+25b^2)=0

We get rid of parentheses

-b^2-25b^2+48=0

We add all the numbers together, and all the variables

-26b^2+48=0

a = -26; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-26)·48
Δ = 4992
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:

$\sqrt{\Delta}=\sqrt{4992}=\sqrt{64*78}=\sqrt{64}*\sqrt{78}=8\sqrt{78}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{78}}{2*-26}=\frac{0-8\sqrt{78}}{-52}
=-\frac{8\sqrt{78}}{-52}
=-\frac{2\sqrt{78}}{-13}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{78}}{2*-26}=\frac{0+8\sqrt{78}}{-52}
=\frac{8\sqrt{78}}{-52}
=\frac{2\sqrt{78}}{-13}
$