(10/m)+(2m)=-20

Solution for (10/m)+(2m)=-20 equation:

(10/m)+(2m)=-20

We move all terms to the left:

(10/m)+(2m)-(-20)=0


Domain of the equation: m)!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+10/m)+2m-(-20)=0

We add all the numbers together, and all the variables

2m+(+10/m)+20=0

We get rid of parentheses

2m+10/m+20=0

We multiply all the terms by the denominator


2m*m+20*m+10=0

We add all the numbers together, and all the variables

20m+2m*m+10=0

Wy multiply elements

2m^2+20m+10=0

a = 2; b = 20; c = +10;
Δ = b2-4ac
Δ = 202-4·2·10
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*2}=\frac{-20-8\sqrt{5}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*2}=\frac{-20+8\sqrt{5}}{4}
$