(3y-2)2=2(y2-2)+3y-8

Solution for (3y-2)2=2(y2-2)+3y-8 equation:

(3y-2)2=2(y2-2)+3y-8

We move all terms to the left:

(3y-2)2-(2(y2-2)+3y-8)=0

We add all the numbers together, and all the variables

-(2(+y^2-2)+3y-8)+(3y-2)2=0

We multiply parentheses

-(2(+y^2-2)+3y-8)+6y-4=0


We calculate terms in parentheses: -(2(+y^2-2)+3y-8), so:

2(+y^2-2)+3y-8

We multiply parentheses

2y^2+3y-4-8

We add all the numbers together, and all the variables

2y^2+3y-12

Back to the equation:

-(2y^2+3y-12)


We add all the numbers together, and all the variables

6y-(2y^2+3y-12)-4=0

We get rid of parentheses

-2y^2+6y-3y+12-4=0

We add all the numbers together, and all the variables

-2y^2+3y+8=0

a = -2; b = 3; c = +8;
Δ = b2-4ac
Δ = 32-4·(-2)·8
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{73}}{2*-2}=\frac{-3-\sqrt{73}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*-2}=\frac{-3+\sqrt{73}}{-4}
$