(1/2y+4)(y-8)=0

Solution for (1/2y+4)(y-8)=0 equation:

(1/2y+4)(y-8)=0


Domain of the equation: 2y+4)(y-8)!=0

y∈R


We multiply parentheses ..

(+y^2-8y+4y-32)=0

We get rid of parentheses

y^2-8y+4y-32=0

We add all the numbers together, and all the variables

y^2-4y-32=0

a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*1}=\frac{-8}{2}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2}
=8
$