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3y^2+16y+16=0
a = 3; b = 16; c = +16;
Δ = b2-4ac
Δ = 162-4·3·16
Δ = 64
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{64}=8$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*3}=\frac{-24}{6} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*3}=\frac{-8}{6} =-1+1/3 $
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