7p^2+16=2152

Solution for 7p^2+16=2152 equation:

7p^2+16=2152

We move all terms to the left:

7p^2+16-(2152)=0

We add all the numbers together, and all the variables

7p^2-2136=0

a = 7; b = 0; c = -2136;
Δ = b2-4ac
Δ = 02-4·7·(-2136)
Δ = 59808
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:

$\sqrt{\Delta}=\sqrt{59808}=\sqrt{16*3738}=\sqrt{16}*\sqrt{3738}=4\sqrt{3738}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3738}}{2*7}=\frac{0-4\sqrt{3738}}{14}
=-\frac{4\sqrt{3738}}{14}
=-\frac{2\sqrt{3738}}{7}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3738}}{2*7}=\frac{0+4\sqrt{3738}}{14}
=\frac{4\sqrt{3738}}{14}
=\frac{2\sqrt{3738}}{7}
$