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2p^2+7p-8=-12
We move all terms to the left:
2p^2+7p-8-(-12)=0
We add all the numbers together, and all the variables
2p^2+7p+4=0
a = 2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·2·4
Δ = 17
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{17}}{2*2}=\frac{-7-\sqrt{17}}{4} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{17}}{2*2}=\frac{-7+\sqrt{17}}{4} $
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