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4y^2+27y+30=0
a = 4; b = 27; c = +30;
Δ = b2-4ac
Δ = 272-4·4·30
Δ = 249
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{249}}{2*4}=\frac{-27-\sqrt{249}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{249}}{2*4}=\frac{-27+\sqrt{249}}{8} $
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