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y+y^2=30
We move all terms to the left:
y+y^2-(30)=0
a = 1; b = 1; c = -30;
Δ = b2-4ac
Δ = 12-4·1·(-30)
Δ = 121
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{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*1}=\frac{-12}{2} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*1}=\frac{10}{2} =5 $
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