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4y^2+13y-12=0
a = 4; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·4·(-12)
Δ = 361
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*4}=\frac{-32}{8} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*4}=\frac{6}{8} =3/4 $
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