Y=3x2+53x

Solution for Y=3x2+53x equation:

=3Y^2+53Y

We move all terms to the left:

-(3Y^2+53Y)=0

We get rid of parentheses

-3Y^2-53Y=0

a = -3; b = -53; c = 0;
Δ = b2-4ac
Δ = -532-4·(-3)·0
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-53}{2*-3}=\frac{0}{-6}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+53}{2*-3}=\frac{106}{-6}
=-17+2/3
$