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3y^2+8y+3=0
a = 3; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·3·3
Δ = 28
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{7}}{2*3}=\frac{-8-2\sqrt{7}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{7}}{2*3}=\frac{-8+2\sqrt{7}}{6} $
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