26^2=10^2+b^2

Solution for 26^2=10^2+b^2 equation:

26^2=10^2+b^2

We move all terms to the left:

26^2-(10^2+b^2)=0

We add all the numbers together, and all the variables

-(10^2+b^2)+676=0

We get rid of parentheses

-b^2+676-10^2=0

We add all the numbers together, and all the variables

-1b^2+576=0

a = -1; b = 0; c = +576;
Δ = b2-4ac
Δ = 02-4·(-1)·576
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-1}=\frac{-48}{-2}
=+24
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-1}=\frac{48}{-2}
=-24
$