1/2x-4=8x-3

Solution for 1/2x-4=8x-3 equation:

1/2x-4=8x-3

We move all terms to the left:

1/2x-4-(8x-3)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We get rid of parentheses

1/2x-8x+3-4=0

We multiply all the terms by the denominator


-8x*2x+3*2x-4*2x+1=0

Wy multiply elements

-16x^2+6x-8x+1=0

We add all the numbers together, and all the variables

-16x^2-2x+1=0

a = -16; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-16)·1
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*-16}=\frac{2-2\sqrt{17}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*-16}=\frac{2+2\sqrt{17}}{-32}
$