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y^2-98y+2376=0
a = 1; b = -98; c = +2376;
Δ = b2-4ac
Δ = -982-4·1·2376
Δ = 100
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-10}{2*1}=\frac{88}{2} =44 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+10}{2*1}=\frac{108}{2} =54 $
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