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4y^2+23y-33=0
a = 4; b = 23; c = -33;
Δ = b2-4ac
Δ = 232-4·4·(-33)
Δ = 1057
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{-(23)-\sqrt{1057}}{2*4}=\frac{-23-\sqrt{1057}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{1057}}{2*4}=\frac{-23+\sqrt{1057}}{8} $
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