d^2-(3/4)d+(1/8)=0

Solution for d^2-(3/4)d+(1/8)=0 equation:

d^2-(3/4)d+(1/8)=0


Domain of the equation: 4)d!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

d^2-(+3/4)d+(+1/8)=0

We multiply parentheses

d^2-3d^2+(+1/8)=0

We get rid of parentheses

d^2-3d^2+1/8=0

We multiply all the terms by the denominator


d^2*8-3d^2*8+1=0

Wy multiply elements

8d^2-24d^2+1=0

We add all the numbers together, and all the variables

-16d^2+1=0

a = -16; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-16)·1
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-16}=\frac{-8}{-32}
=1/4
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-16}=\frac{8}{-32}
=-1/4
$