40=-1/5q+10q

Solution for 40=-1/5q+10q equation:

40=-1/5q+10q

We move all terms to the left:

40-(-1/5q+10q)=0


Domain of the equation: 5q+10q)!=0

q∈R


We add all the numbers together, and all the variables

-(+10q-1/5q)+40=0

We get rid of parentheses

-10q+1/5q+40=0

We multiply all the terms by the denominator


-10q*5q+40*5q+1=0

Wy multiply elements

-50q^2+200q+1=0

a = -50; b = 200; c = +1;
Δ = b2-4ac
Δ = 2002-4·(-50)·1
Δ = 40200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40200}=\sqrt{100*402}=\sqrt{100}*\sqrt{402}=10\sqrt{402}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-10\sqrt{402}}{2*-50}=\frac{-200-10\sqrt{402}}{-100}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+10\sqrt{402}}{2*-50}=\frac{-200+10\sqrt{402}}{-100}
$