x^2-9/5x+4=x^2-9/4x+5

Solution for x^2-9/5x+4=x^2-9/4x+5 equation:

x^2-9/5x+4=x^2-9/4x+5

We move all terms to the left:

x^2-9/5x+4-(x^2-9/4x+5)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 4x+5)!=0

x∈R


We get rid of parentheses

x^2-x^2-9/5x+9/4x-5+4=0

We calculate fractions


x^2-x^2+(-36x)/20x^2+45x/20x^2-5+4=0

We add all the numbers together, and all the variables

(-36x)/20x^2+45x/20x^2-1=0

We multiply all the terms by the denominator


(-36x)+45x-1*20x^2=0

We add all the numbers together, and all the variables

45x+(-36x)-1*20x^2=0

Wy multiply elements

-20x^2+45x+(-36x)=0

We get rid of parentheses

-20x^2+45x-36x=0

We add all the numbers together, and all the variables

-20x^2+9x=0

a = -20; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-20)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-20}=\frac{-18}{-40}
=9/20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-20}=\frac{0}{-40}
=0
$