p^2+40=14p^2

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

p^2+40=14p^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

-13p^2+40=0

a = -13; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-13)·40
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{130}}{2*-13}=\frac{0-4\sqrt{130}}{-26}
=-\frac{4\sqrt{130}}{-26}
=-\frac{2\sqrt{130}}{-13}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{130}}{2*-13}=\frac{0+4\sqrt{130}}{-26}
=\frac{4\sqrt{130}}{-26}
=\frac{2\sqrt{130}}{-13}
$