(2/3)d+8=48

Solution for (2/3)d+8=48 equation:

(2/3)d+8=48

We move all terms to the left:

(2/3)d+8-(48)=0


Domain of the equation: 3)d!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

(+2/3)d+8-48=0

We add all the numbers together, and all the variables

(+2/3)d-40=0

We multiply parentheses

2d^2-40=0

a = 2; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·2·(-40)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*2}=\frac{0-8\sqrt{5}}{4}
=-\frac{8\sqrt{5}}{4}
=-2\sqrt{5}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*2}=\frac{0+8\sqrt{5}}{4}
=\frac{8\sqrt{5}}{4}
=2\sqrt{5}
$