p^2+10p=35

Solution for p^2+10p=35 equation:

p^2+10p=35

We move all terms to the left:

p^2+10p-(35)=0

a = 1; b = 10; c = -35;
Δ = b2-4ac
Δ = 102-4·1·(-35)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{15}}{2*1}=\frac{-10-4\sqrt{15}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{15}}{2*1}=\frac{-10+4\sqrt{15}}{2}
$