Y=3x^2+2x-13

Solution for Y=3x^2+2x-13 equation:

=3Y^2+2Y-13

We move all terms to the left:

-(3Y^2+2Y-13)=0

We get rid of parentheses

-3Y^2-2Y+13=0

a = -3; b = -2; c = +13;
Δ = b2-4ac
Δ = -22-4·(-3)·13
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{10}}{2*-3}=\frac{2-4\sqrt{10}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{10}}{2*-3}=\frac{2+4\sqrt{10}}{-6}
$