(2+3/4)w+4=48

Solution for (2+3/4)w+4=48 equation:

(2+3/4)w+4=48

We move all terms to the left:

(2+3/4)w+4-(48)=0


Domain of the equation: 4)w!=0

w!=0/1

w!=0

w∈R


We add all the numbers together, and all the variables

(3/4+2)w+4-48=0

We add all the numbers together, and all the variables

(3/4+2)w-44=0

We multiply parentheses

3w^2+2w-44=0

a = 3; b = 2; c = -44;
Δ = b2-4ac
Δ = 22-4·3·(-44)
Δ = 532
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:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{133}}{2*3}=\frac{-2-2\sqrt{133}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{133}}{2*3}=\frac{-2+2\sqrt{133}}{6}
$