(3/4)x+2-3x=20

Solution for (3/4)x+2-3x=20 equation:

(3/4)x+2-3x=20

We move all terms to the left:

(3/4)x+2-3x-(20)=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-3x+2-20=0

We add all the numbers together, and all the variables

-3x+(+3/4)x-18=0

We multiply parentheses

3x^2-3x-18=0

a = 3; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·3·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*3}=\frac{18}{6}
=3
$