25+b2=169

Solution for 25+b2=169 equation:

25+b2=169

We move all terms to the left:

25+b2-(169)=0

We add all the numbers together, and all the variables

b^2-144=0

a = 1; b = 0; c = -144;
Δ = b2-4ac
Δ = 02-4·1·(-144)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*1}=\frac{-24}{2}
=-12
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*1}=\frac{24}{2}
=12
$