y=(-3/2)+4,y=(-3/2)-3

Solution for y=(-3/2)+4,y=(-3/2)-3 equation:

y=(-3/2)+4.y=(-3/2)-3

We move all terms to the left:

y-((-3/2)+4.y)=0


Domain of the equation: 2)+4.y)!=0

y!=0/1

y!=0

y∈R


We multiply all the terms by the denominator


y*2)+4.y)-((-3=0

Wy multiply elements

2y^2-3=0

a = 2; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·2·(-3)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*2}=\frac{0-2\sqrt{6}}{4}
=-\frac{2\sqrt{6}}{4}
=-\frac{\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*2}=\frac{0+2\sqrt{6}}{4}
=\frac{2\sqrt{6}}{4}
=\frac{\sqrt{6}}{2}
$