2y^2+3y-2=(y+2)

Solution for 2y^2+3y-2=(y+2) equation:

2y^2+3y-2=(y+2)

We move all terms to the left:

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


We calculate terms in parentheses: -((y+2)), so:

(y+2)

We get rid of parentheses

y+2

Back to the equation:

-(y+2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2+2y-4=0

a = 2; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·2·(-4)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*2}=\frac{-8}{4}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*2}=\frac{4}{4}
=1
$