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y^2+2y-50=0
a = 1; b = 2; c = -50;
Δ = b2-4ac
Δ = 22-4·1·(-50)
Δ = 204
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{51}}{2*1}=\frac{-2-2\sqrt{51}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{51}}{2*1}=\frac{-2+2\sqrt{51}}{2} $
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