8(2y-5)y=1/2

Solution for 8(2y-5)y=1/2 equation:

8(2y-5)y=1/2

We move all terms to the left:

8(2y-5)y-(1/2)=0

We add all the numbers together, and all the variables

8(2y-5)y-(+1/2)=0

We multiply parentheses

16y^2-40y-(+1/2)=0

We get rid of parentheses

16y^2-40y-1/2=0

We multiply all the terms by the denominator


16y^2*2-40y*2-1=0

Wy multiply elements

32y^2-80y-1=0

a = 32; b = -80; c = -1;
Δ = b2-4ac
Δ = -802-4·32·(-1)
Δ = 6528
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:

$\sqrt{\Delta}=\sqrt{6528}=\sqrt{64*102}=\sqrt{64}*\sqrt{102}=8\sqrt{102}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-8\sqrt{102}}{2*32}=\frac{80-8\sqrt{102}}{64}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+8\sqrt{102}}{2*32}=\frac{80+8\sqrt{102}}{64}
$