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

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

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

We move all terms to the left:

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


Domain of the equation: 5w)!=0

w!=0/1

w!=0

w∈R



Domain of the equation: 2w)!=0

w!=0/1

w!=0

w∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


14w/10w^2+15w/10w^2-1=0

We multiply all the terms by the denominator


14w+15w-1*10w^2=0

We add all the numbers together, and all the variables

29w-1*10w^2=0

Wy multiply elements

-10w^2+29w=0

a = -10; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·(-10)·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*-10}=\frac{-58}{-20}
=2+9/10
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*-10}=\frac{0}{-20}
=0
$