p+p2=1260

Solution for p+p2=1260 equation:

p+p2=1260

We move all terms to the left:

p+p2-(1260)=0

We add all the numbers together, and all the variables

p^2+p-1260=0

a = 1; b = 1; c = -1260;
Δ = b2-4ac
Δ = 12-4·1·(-1260)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-71}{2*1}=\frac{-72}{2}
=-36
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+71}{2*1}=\frac{70}{2}
=35
$