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2y^2-10y-325=0
a = 2; b = -10; c = -325;
Δ = b2-4ac
Δ = -102-4·2·(-325)
Δ = 2700
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-30\sqrt{3}}{2*2}=\frac{10-30\sqrt{3}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+30\sqrt{3}}{2*2}=\frac{10+30\sqrt{3}}{4} $
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