(1/8)d-4+(3/8)d-4=5

Solution for (1/8)d-4+(3/8)d-4=5 equation:

(1/8)d-4+(3/8)d-4=5

We move all terms to the left:

(1/8)d-4+(3/8)d-4-(5)=0


Domain of the equation: 8)d!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

(+1/8)d+(+3/8)d-4-4-5=0

We add all the numbers together, and all the variables

(+1/8)d+(+3/8)d-13=0

We multiply parentheses

d^2+3d^2-13=0

We add all the numbers together, and all the variables

4d^2-13=0

a = 4; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·4·(-13)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*4}=\frac{0-4\sqrt{13}}{8}
=-\frac{4\sqrt{13}}{8}
=-\frac{\sqrt{13}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*4}=\frac{0+4\sqrt{13}}{8}
=\frac{4\sqrt{13}}{8}
=\frac{\sqrt{13}}{2}
$