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3p^2+23p+14=0
a = 3; b = 23; c = +14;
Δ = b2-4ac
Δ = 232-4·3·14
Δ = 361
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{361}=19$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*3}=\frac{-42}{6} =-7 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*3}=\frac{-4}{6} =-2/3 $
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