7p^2+16=2161

Solution for 7p^2+16=2161 equation:

7p^2+16=2161

We move all terms to the left:

7p^2+16-(2161)=0

We add all the numbers together, and all the variables

7p^2-2145=0

a = 7; b = 0; c = -2145;
Δ = b2-4ac
Δ = 02-4·7·(-2145)
Δ = 60060
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:

$\sqrt{\Delta}=\sqrt{60060}=\sqrt{4*15015}=\sqrt{4}*\sqrt{15015}=2\sqrt{15015}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15015}}{2*7}=\frac{0-2\sqrt{15015}}{14}
=-\frac{2\sqrt{15015}}{14}
=-\frac{\sqrt{15015}}{7}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15015}}{2*7}=\frac{0+2\sqrt{15015}}{14}
=\frac{2\sqrt{15015}}{14}
=\frac{\sqrt{15015}}{7}
$