Y=5x^2/3+3x

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

=5Y^2/3+3Y

We move all terms to the left:

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


Domain of the equation: 3+3Y)!=0

We move all terms containing Y to the left, all other terms to the right

3Y)!=-3

Y!=-3/1

Y!=-3

Y∈R


We get rid of parentheses

-5Y^2/3-3Y=0

We multiply all the terms by the denominator


-5Y^2-3Y*3=0

Wy multiply elements

-5Y^2-9Y=0

a = -5; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-5)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-5}=\frac{0}{-10}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-5}=\frac{18}{-10}
=-1+4/5
$