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-4w^2+3w+6=0
a = -4; b = 3; c = +6;
Δ = b2-4ac
Δ = 32-4·(-4)·6
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{105}}{2*-4}=\frac{-3-\sqrt{105}}{-8} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{105}}{2*-4}=\frac{-3+\sqrt{105}}{-8} $
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