3p^2+10p^2-8^2=0

Solution for 3p^2+10p^2-8^2=0 equation:

3p^2+10p^2-8^2=0

We add all the numbers together, and all the variables

13p^2-64=0

a = 13; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·13·(-64)
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{13}}{2*13}=\frac{0-16\sqrt{13}}{26}
=-\frac{16\sqrt{13}}{26}
=-\frac{8\sqrt{13}}{13}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{13}}{2*13}=\frac{0+16\sqrt{13}}{26}
=\frac{16\sqrt{13}}{26}
=\frac{8\sqrt{13}}{13}
$