25+b^2=81

Solution for 25+b^2=81 equation:

25+b^2=81

We move all terms to the left:

25+b^2-(81)=0

We add all the numbers together, and all the variables

b^2-56=0

a = 1; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·1·(-56)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*1}=\frac{0-4\sqrt{14}}{2}
=-\frac{4\sqrt{14}}{2}
=-2\sqrt{14}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*1}=\frac{0+4\sqrt{14}}{2}
=\frac{4\sqrt{14}}{2}
=2\sqrt{14}
$