1/2d-3+1/2d=d-3

Solution for 1/2d-3+1/2d=d-3 equation:

1/2d-3+1/2d=d-3

We move all terms to the left:

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


Domain of the equation: 2d!=0

d!=0/2

d!=0

d∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-d*2d+3*2d-3*2d+2=0

Wy multiply elements

-2d^2+6d-6d+2=0

We add all the numbers together, and all the variables

-2d^2+2=0

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

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