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4y^2-40y+75=0
a = 4; b = -40; c = +75;
Δ = b2-4ac
Δ = -402-4·4·75
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*4}=\frac{20}{8} =2+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*4}=\frac{60}{8} =7+1/2 $
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