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y^2-44y+96=0
a = 1; b = -44; c = +96;
Δ = b2-4ac
Δ = -442-4·1·96
Δ = 1552
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{97}}{2*1}=\frac{44-4\sqrt{97}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{97}}{2*1}=\frac{44+4\sqrt{97}}{2} $
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