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3p^2+10p-8=0
a = 3; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·3·(-8)
Δ = 196
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{196}=14$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*3}=\frac{-24}{6} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*3}=\frac{4}{6} =2/3 $
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