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p^2=4-3p
We move all terms to the left:
p^2-(4-3p)=0
We add all the numbers together, and all the variables
p^2-(-3p+4)=0
We get rid of parentheses
p^2+3p-4=0
a = 1; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·1·(-4)
Δ = 25
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{25}=5$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*1}=\frac{-8}{2} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*1}=\frac{2}{2} =1 $
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