p^2=7p+10

Solution for p^2=7p+10 equation:

p^2=7p+10

We move all terms to the left:

p^2-(7p+10)=0

We get rid of parentheses

p^2-7p-10=0

a = 1; b = -7; c = -10;
Δ = b2-4ac
Δ = -72-4·1·(-10)
Δ = 89
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{-(-7)-\sqrt{89}}{2*1}=\frac{7-\sqrt{89}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{89}}{2*1}=\frac{7+\sqrt{89}}{2}
$