p^2+16p+7=0

Solution for p^2+16p+7=0 equation:

p^2+16p+7=0

a = 1; b = 16; c = +7;
Δ = b2-4ac
Δ = 162-4·1·7
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{57}}{2*1}=\frac{-16-2\sqrt{57}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{57}}{2*1}=\frac{-16+2\sqrt{57}}{2}
$