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a^2+32^2=37^2
We move all terms to the left:
a^2+32^2-(37^2)=0
We add all the numbers together, and all the variables
a^2-345=0
a = 1; b = 0; c = -345;
Δ = b2-4ac
Δ = 02-4·1·(-345)
Δ = 1380
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1380}=\sqrt{4*345}=\sqrt{4}*\sqrt{345}=2\sqrt{345}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{345}}{2*1}=\frac{0-2\sqrt{345}}{2} =-\frac{2\sqrt{345}}{2} =-\sqrt{345} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{345}}{2*1}=\frac{0+2\sqrt{345}}{2} =\frac{2\sqrt{345}}{2} =\sqrt{345} $
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