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3y^2+14y+15=0
a = 3; b = 14; c = +15;
Δ = b2-4ac
Δ = 142-4·3·15
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*3}=\frac{-18}{6} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*3}=\frac{-10}{6} =-1+2/3 $
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