7/2w+3/5w=1

Solution for 7/2w+3/5w=1 equation:

7/2w+3/5w=1

We move all terms to the left:

7/2w+3/5w-(1)=0


Domain of the equation: 2w!=0

w!=0/2

w!=0

w∈R



Domain of the equation: 5w!=0

w!=0/5

w!=0

w∈R


We calculate fractions


35w/10w^2+6w/10w^2-1=0

We multiply all the terms by the denominator


35w+6w-1*10w^2=0

We add all the numbers together, and all the variables

41w-1*10w^2=0

Wy multiply elements

-10w^2+41w=0

a = -10; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·(-10)·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*-10}=\frac{-82}{-20}
=4+1/10
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*-10}=\frac{0}{-20}
=0
$