(P-4)=p^2+3p-5

Solution for (P-4)=p^2+3p-5 equation:

(-4)=P^2+3P-5

We move all terms to the left:

(-4)-(P^2+3P-5)=0

We add all the numbers together, and all the variables

-(P^2+3P-5)-4=0

We get rid of parentheses

-P^2-3P+5-4=0

We add all the numbers together, and all the variables

-1P^2-3P+1=0

a = -1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-1)·1
Δ = 13
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{-(-3)-\sqrt{13}}{2*-1}=\frac{3-\sqrt{13}}{-2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*-1}=\frac{3+\sqrt{13}}{-2}
$