8x+((1-4x)/8))=7

Solution for 8x+((1-4x)/8))=7 equation:

8x+((1-4x)/8))=7

We move all terms to the left:

8x+((1-4x)/8))-(7)=0

We add all the numbers together, and all the variables

8x+((-4x+1)/8))-7=0

We multiply all the terms by the denominator


8x*8))-7+((-4x+1)=0

We add all the numbers together, and all the variables

-4x+8x*8))-7+((=0

Wy multiply elements

64x^2-4x=0

a = 64; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·64·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*64}=\frac{0}{128}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*64}=\frac{8}{128}
=1/16
$