Y=3x^2+8x+1

Solution for Y=3x^2+8x+1 equation:

=3Y^2+8Y+1

We move all terms to the left:

-(3Y^2+8Y+1)=0

We get rid of parentheses

-3Y^2-8Y-1=0

a = -3; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·(-3)·(-1)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{13}}{2*-3}=\frac{8-2\sqrt{13}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{13}}{2*-3}=\frac{8+2\sqrt{13}}{-6}
$