Y=3x^2+4x-5

Solution for Y=3x^2+4x-5 equation:

=3Y^2+4Y-5

We move all terms to the left:

-(3Y^2+4Y-5)=0

We get rid of parentheses

-3Y^2-4Y+5=0

a = -3; b = -4; c = +5;
Δ = b2-4ac
Δ = -42-4·(-3)·5
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{19}}{2*-3}=\frac{4-2\sqrt{19}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{19}}{2*-3}=\frac{4+2\sqrt{19}}{-6}
$