6=(3/5)(10-5x)

Solution for 6=(3/5)(10-5x) equation:

6=(3/5)(10-5x)

We move all terms to the left:

6-((3/5)(10-5x))=0


Domain of the equation: 5)(10-5x))!=0

x∈R


We add all the numbers together, and all the variables

-((+3/5)(-5x+10))+6=0

We multiply parentheses ..

-((-15x^2+3/5*10))+6=0

We multiply all the terms by the denominator


-((-15x^2+3+6*5*10))=0


We calculate terms in parentheses: -((-15x^2+3+6*5*10)), so:

(-15x^2+3+6*5*10)

We get rid of parentheses

-15x^2+3+6*5*10

We add all the numbers together, and all the variables

-15x^2+303

Back to the equation:

-(-15x^2+303)


We get rid of parentheses

15x^2-303=0

a = 15; b = 0; c = -303;
Δ = b2-4ac
Δ = 02-4·15·(-303)
Δ = 18180
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:

$\sqrt{\Delta}=\sqrt{18180}=\sqrt{36*505}=\sqrt{36}*\sqrt{505}=6\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{505}}{2*15}=\frac{0-6\sqrt{505}}{30}
=-\frac{6\sqrt{505}}{30}
=-\frac{\sqrt{505}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{505}}{2*15}=\frac{0+6\sqrt{505}}{30}
=\frac{6\sqrt{505}}{30}
=\frac{\sqrt{505}}{5}
$