5p^2-27p-40=2p^2

Solution for 5p^2-27p-40=2p^2 equation:

5p^2-27p-40=2p^2

We move all terms to the left:

5p^2-27p-40-(2p^2)=0

determiningTheFunctionDomain
5p^2-2p^2-27p-40=0

We add all the numbers together, and all the variables

3p^2-27p-40=0

a = 3; b = -27; c = -40;
Δ = b2-4ac
Δ = -272-4·3·(-40)
Δ = 1209
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{-(-27)-\sqrt{1209}}{2*3}=\frac{27-\sqrt{1209}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{1209}}{2*3}=\frac{27+\sqrt{1209}}{6}
$