p^2+7p+10=5

Solution for p^2+7p+10=5 equation:

p^2+7p+10=5

We move all terms to the left:

p^2+7p+10-(5)=0

We add all the numbers together, and all the variables

p^2+7p+5=0

a = 1; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·1·5
Δ = 29
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{29}}{2*1}=\frac{-7-\sqrt{29}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{29}}{2*1}=\frac{-7+\sqrt{29}}{2}
$