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4y^2+20y-11=0
a = 4; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·4·(-11)
Δ = 576
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{576}=24$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*4}=\frac{-44}{8} =-5+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*4}=\frac{4}{8} =1/2 $
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