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3p^2+4=-7p
We move all terms to the left:
3p^2+4-(-7p)=0
We get rid of parentheses
3p^2+7p+4=0
a = 3; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·3·4
Δ = 1
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{1}=1$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*3}=\frac{-8}{6} =-1+1/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*3}=\frac{-6}{6} =-1 $
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