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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

-8(y-2)

We multiply parentheses

-8y+16

Back to the equation:

-(-8y+16)


We get rid of parentheses

2y^2+6y+8y-16=0

We add all the numbers together, and all the variables

2y^2+14y-16=0

a = 2; b = 14; c = -16;
Δ = b2-4ac
Δ = 142-4·2·(-16)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*2}=\frac{-32}{4}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*2}=\frac{4}{4}
=1
$