3y/2+y(y=3)

Solution for 3y/2+y(y=3) equation:

3y/2+y(y=3)

We move all terms to the left:

3y/2+y(y-(3))=0

We multiply parentheses

y^2+3y/2-3y=0

We multiply all the terms by the denominator


y^2*2+3y-3y*2=0

Wy multiply elements

2y^2+3y-6y=0

We add all the numbers together, and all the variables

2y^2-3y=0

a = 2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*2}=\frac{0}{4}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*2}=\frac{6}{4}
=1+1/2
$