9/5x+42=3x

Solution for 9/5x+42=3x equation:

9/5x+42=3x

We move all terms to the left:

9/5x+42-(3x)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We add all the numbers together, and all the variables

-3x+9/5x+42=0

We multiply all the terms by the denominator


-3x*5x+42*5x+9=0

Wy multiply elements

-15x^2+210x+9=0

a = -15; b = 210; c = +9;
Δ = b2-4ac
Δ = 2102-4·(-15)·9
Δ = 44640
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:

$\sqrt{\Delta}=\sqrt{44640}=\sqrt{144*310}=\sqrt{144}*\sqrt{310}=12\sqrt{310}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(210)-12\sqrt{310}}{2*-15}=\frac{-210-12\sqrt{310}}{-30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(210)+12\sqrt{310}}{2*-15}=\frac{-210+12\sqrt{310}}{-30}
$