90+(2/5x)+(3/4x-2)=180

Solution for 90+(2/5x)+(3/4x-2)=180 equation:

90+(2/5x)+(3/4x-2)=180

We move all terms to the left:

90+(2/5x)+(3/4x-2)-(180)=0


Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4x-2)!=0

x∈R


We add all the numbers together, and all the variables

(+2/5x)+(3/4x-2)+90-180=0

We add all the numbers together, and all the variables

(+2/5x)+(3/4x-2)-90=0

We get rid of parentheses

2/5x+3/4x-2-90=0

We calculate fractions


8x/20x^2+15x/20x^2-2-90=0

We add all the numbers together, and all the variables

8x/20x^2+15x/20x^2-92=0

We multiply all the terms by the denominator


8x+15x-92*20x^2=0

We add all the numbers together, and all the variables

23x-92*20x^2=0

Wy multiply elements

-1840x^2+23x=0

a = -1840; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·(-1840)·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*-1840}=\frac{-46}{-3680}
=1/80
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*-1840}=\frac{0}{-3680}
=0
$