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4y^2+6y-28=0
a = 4; b = 6; c = -28;
Δ = b2-4ac
Δ = 62-4·4·(-28)
Δ = 484
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{484}=22$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-22}{2*4}=\frac{-28}{8} =-3+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*4}=\frac{16}{8} =2 $
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