P(x)=3X^2+X-8

Solution for P(x)=3X^2+X-8 equation:

(P)=3P^2+P-8

We move all terms to the left:

(P)-(3P^2+P-8)=0

We get rid of parentheses

-3P^2+P-P+8=0

We add all the numbers together, and all the variables

-3P^2+8=0

a = -3; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-3)·8
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-3}=\frac{0-4\sqrt{6}}{-6}
=-\frac{4\sqrt{6}}{-6}
=-\frac{2\sqrt{6}}{-3}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-3}=\frac{0+4\sqrt{6}}{-6}
=\frac{4\sqrt{6}}{-6}
=\frac{2\sqrt{6}}{-3}
$