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2y^2+3y-12=0
a = 2; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·2·(-12)
Δ = 105
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{105}}{2*2}=\frac{-3-\sqrt{105}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{105}}{2*2}=\frac{-3+\sqrt{105}}{4} $
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