(p)+(p2)+(p2+10)=70

Solution for (p)+(p2)+(p2+10)=70 equation:

(p)+(p2)+(p2+10)=70

We move all terms to the left:

(p)+(p2)+(p2+10)-(70)=0

We add all the numbers together, and all the variables

(+p^2+10)+p+p2-70=0

We add all the numbers together, and all the variables

p^2+(+p^2+10)+p-70=0

We get rid of parentheses

p^2+p^2+p+10-70=0

We add all the numbers together, and all the variables

2p^2+p-60=0

a = 2; b = 1; c = -60;
Δ = b2-4ac
Δ = 12-4·2·(-60)
Δ = 481
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{-(1)-\sqrt{481}}{2*2}=\frac{-1-\sqrt{481}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{481}}{2*2}=\frac{-1+\sqrt{481}}{4}
$