(3/4)w-(1/2)w-4=12

Solution for (3/4)w-(1/2)w-4=12 equation:

(3/4)w-(1/2)w-4=12

We move all terms to the left:

(3/4)w-(1/2)w-4-(12)=0


Domain of the equation: 4)w!=0

w!=0/1

w!=0

w∈R



Domain of the equation: 2)w!=0

w!=0/1

w!=0

w∈R


We add all the numbers together, and all the variables

(+3/4)w-(+1/2)w-4-12=0

We add all the numbers together, and all the variables

(+3/4)w-(+1/2)w-16=0

We multiply parentheses

3w^2-w^2-16=0

We add all the numbers together, and all the variables

2w^2-16=0

a = 2; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·2·(-16)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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