10/3x-36=2x+96

Solution for 10/3x-36=2x+96 equation:

10/3x-36=2x+96

We move all terms to the left:

10/3x-36-(2x+96)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

10/3x-2x-96-36=0

We multiply all the terms by the denominator


-2x*3x-96*3x-36*3x+10=0

Wy multiply elements

-6x^2-288x-108x+10=0

We add all the numbers together, and all the variables

-6x^2-396x+10=0

a = -6; b = -396; c = +10;
Δ = b2-4ac
Δ = -3962-4·(-6)·10
Δ = 157056
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:

$\sqrt{\Delta}=\sqrt{157056}=\sqrt{64*2454}=\sqrt{64}*\sqrt{2454}=8\sqrt{2454}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-396)-8\sqrt{2454}}{2*-6}=\frac{396-8\sqrt{2454}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-396)+8\sqrt{2454}}{2*-6}=\frac{396+8\sqrt{2454}}{-12}
$