(3/4)x-4=40-2x

Solution for (3/4)x-4=40-2x equation:

(3/4)x-4=40-2x

We move all terms to the left:

(3/4)x-4-(40-2x)=0


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)x-(-2x+40)-4=0

We multiply parentheses

3x^2-(-2x+40)-4=0

We get rid of parentheses

3x^2+2x-40-4=0

We add all the numbers together, and all the variables

3x^2+2x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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