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3^2+6^2=c^2
We move all terms to the left:
3^2+6^2-(c^2)=0
We add all the numbers together, and all the variables
-1c^2+45=0
a = -1; b = 0; c = +45;
Δ = b2-4ac
Δ = 02-4·(-1)·45
Δ = 180
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-1}=\frac{0-6\sqrt{5}}{-2} =-\frac{6\sqrt{5}}{-2} =-\frac{3\sqrt{5}}{-1} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-1}=\frac{0+6\sqrt{5}}{-2} =\frac{6\sqrt{5}}{-2} =\frac{3\sqrt{5}}{-1} $
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