2x=9/5*x+32

Solution for 2x=9/5*x+32 equation:

2x=9/5x+32

We move all terms to the left:

2x-(9/5x+32)=0


Domain of the equation: 5x+32)!=0

x∈R


We get rid of parentheses

2x-9/5x-32=0

We multiply all the terms by the denominator


2x*5x-32*5x-9=0

Wy multiply elements

10x^2-160x-9=0

a = 10; b = -160; c = -9;
Δ = b2-4ac
Δ = -1602-4·10·(-9)
Δ = 25960
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:

$\sqrt{\Delta}=\sqrt{25960}=\sqrt{4*6490}=\sqrt{4}*\sqrt{6490}=2\sqrt{6490}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-2\sqrt{6490}}{2*10}=\frac{160-2\sqrt{6490}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+2\sqrt{6490}}{2*10}=\frac{160+2\sqrt{6490}}{20}
$