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a^2+a^2=180^2
We move all terms to the left:
a^2+a^2-(180^2)=0
We add all the numbers together, and all the variables
2a^2-32400=0
a = 2; b = 0; c = -32400;
Δ = b2-4ac
Δ = 02-4·2·(-32400)
Δ = 259200
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{259200}=\sqrt{129600*2}=\sqrt{129600}*\sqrt{2}=360\sqrt{2}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-360\sqrt{2}}{2*2}=\frac{0-360\sqrt{2}}{4} =-\frac{360\sqrt{2}}{4} =-90\sqrt{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+360\sqrt{2}}{2*2}=\frac{0+360\sqrt{2}}{4} =\frac{360\sqrt{2}}{4} =90\sqrt{2} $
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