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y2=325
We move all terms to the left:
y2-(325)=0
We add all the numbers together, and all the variables
y^2-325=0
a = 1; b = 0; c = -325;
Δ = b2-4ac
Δ = 02-4·1·(-325)
Δ = 1300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*1}=\frac{0-10\sqrt{13}}{2} =-\frac{10\sqrt{13}}{2} =-5\sqrt{13} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*1}=\frac{0+10\sqrt{13}}{2} =\frac{10\sqrt{13}}{2} =5\sqrt{13} $
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