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y^2+5y=10
We move all terms to the left:
y^2+5y-(10)=0
a = 1; b = 5; c = -10;
Δ = b2-4ac
Δ = 52-4·1·(-10)
Δ = 65
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{65}}{2*1}=\frac{-5-\sqrt{65}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{65}}{2*1}=\frac{-5+\sqrt{65}}{2} $
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