12p^2+13=3001

Solution for 12p^2+13=3001 equation:

12p^2+13=3001

We move all terms to the left:

12p^2+13-(3001)=0

We add all the numbers together, and all the variables

12p^2-2988=0

a = 12; b = 0; c = -2988;
Δ = b2-4ac
Δ = 02-4·12·(-2988)
Δ = 143424
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:

$\sqrt{\Delta}=\sqrt{143424}=\sqrt{576*249}=\sqrt{576}*\sqrt{249}=24\sqrt{249}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{249}}{2*12}=\frac{0-24\sqrt{249}}{24}
=-\frac{24\sqrt{249}}{24}
=-\sqrt{249}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{249}}{2*12}=\frac{0+24\sqrt{249}}{24}
=\frac{24\sqrt{249}}{24}
=\sqrt{249}
$