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4y^2-40y+36=0
a = 4; b = -40; c = +36;
Δ = b2-4ac
Δ = -402-4·4·36
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-32}{2*4}=\frac{8}{8} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+32}{2*4}=\frac{72}{8} =9 $
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