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a^2+a^2=20^2
We move all terms to the left:
a^2+a^2-(20^2)=0
We add all the numbers together, and all the variables
2a^2-400=0
a = 2; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·2·(-400)
Δ = 3200
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*2}=\frac{0-40\sqrt{2}}{4} =-\frac{40\sqrt{2}}{4} =-10\sqrt{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*2}=\frac{0+40\sqrt{2}}{4} =\frac{40\sqrt{2}}{4} =10\sqrt{2} $
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