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4y^2+3y-4=0
a = 4; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·4·(-4)
Δ = 73
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{-(3)-\sqrt{73}}{2*4}=\frac{-3-\sqrt{73}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*4}=\frac{-3+\sqrt{73}}{8} $
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