-(3/5)x+(3/5)=(18/10)

Solution for -(3/5)x+(3/5)=(18/10) equation:

-(3/5)x+(3/5)=(18/10)

We move all terms to the left:

-(3/5)x+(3/5)-((18/10))=0


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-(+3/5)x+(+3/5)-((+18/10))=0

We multiply parentheses

-3x^2+(+3/5)-((+18/10))=0

We get rid of parentheses

-3x^2+3/5-((+18/10))=0

We calculate fractions


-3x^2+()/()+()/()=0

We add all the numbers together, and all the variables

-3x^2+2=0

a = -3; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-3)·2
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-3}=\frac{0-2\sqrt{6}}{-6}
=-\frac{2\sqrt{6}}{-6}
=-\frac{\sqrt{6}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-3}=\frac{0+2\sqrt{6}}{-6}
=\frac{2\sqrt{6}}{-6}
=\frac{\sqrt{6}}{-3}
$