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6^2+4^2=x^2
We move all terms to the left:
6^2+4^2-(x^2)=0
We add all the numbers together, and all the variables
-1x^2+52=0
a = -1; b = 0; c = +52;
Δ = b2-4ac
Δ = 02-4·(-1)·52
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*-1}=\frac{0-4\sqrt{13}}{-2} =-\frac{4\sqrt{13}}{-2} =-\frac{2\sqrt{13}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*-1}=\frac{0+4\sqrt{13}}{-2} =\frac{4\sqrt{13}}{-2} =\frac{2\sqrt{13}}{-1} $
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