121+25=c^2

Solution for 121+25=c^2 equation:

121+25=c^2

We move all terms to the left:

121+25-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+146=0

a = -1; b = 0; c = +146;
Δ = b2-4ac
Δ = 02-4·(-1)·146
Δ = 584
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{584}=\sqrt{4*146}=\sqrt{4}*\sqrt{146}=2\sqrt{146}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{146}}{2*-1}=\frac{0-2\sqrt{146}}{-2}
=-\frac{2\sqrt{146}}{-2}
=-\frac{\sqrt{146}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{146}}{2*-1}=\frac{0+2\sqrt{146}}{-2}
=\frac{2\sqrt{146}}{-2}
=\frac{\sqrt{146}}{-1}
$