p^2+14p+40=6

Solution for p^2+14p+40=6 equation:

p^2+14p+40=6

We move all terms to the left:

p^2+14p+40-(6)=0

We add all the numbers together, and all the variables

p^2+14p+34=0

a = 1; b = 14; c = +34;
Δ = b2-4ac
Δ = 142-4·1·34
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{15}}{2*1}=\frac{-14-2\sqrt{15}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{15}}{2*1}=\frac{-14+2\sqrt{15}}{2}
$