122+142=c2

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Solution for 122+142=c2 equation:



122+142=c2
We move all terms to the left:
122+142-(c2)=0
We add all the numbers together, and all the variables
-1c^2+264=0
a = -1; b = 0; c = +264;
Δ = b2-4ac
Δ = 02-4·(-1)·264
Δ = 1056
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:
$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{66}}{2*-1}=\frac{0-4\sqrt{66}}{-2} =-\frac{4\sqrt{66}}{-2} =-\frac{2\sqrt{66}}{-1} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{66}}{2*-1}=\frac{0+4\sqrt{66}}{-2} =\frac{4\sqrt{66}}{-2} =\frac{2\sqrt{66}}{-1} $

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