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36+5y-y^2=0
We add all the numbers together, and all the variables
-1y^2+5y+36=0
a = -1; b = 5; c = +36;
Δ = b2-4ac
Δ = 52-4·(-1)·36
Δ = 169
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{169}=13$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*-1}=\frac{-18}{-2} =+9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-1}=\frac{8}{-2} =-4 $
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