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3p^2+14p+8=0
a = 3; b = 14; c = +8;
Δ = b2-4ac
Δ = 142-4·3·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*3}=\frac{-24}{6} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*3}=\frac{-4}{6} =-2/3 $
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