1/4(20x+20)-18=-4/5(25x-40)

Solution for 1/4(20x+20)-18=-4/5(25x-40) equation:

1/4(20x+20)-18=-4/5(25x-40)

We move all terms to the left:

1/4(20x+20)-18-(-4/5(25x-40))=0


Domain of the equation: 4(20x+20)!=0

x∈R



Domain of the equation: 5(25x-40))!=0

x∈R


We calculate fractions


(5x2/(4(20x+20)*5(25x-40)))+(-(-16x2)/(4(20x+20)*5(25x-40)))-18=0


We calculate terms in parentheses: +(5x2/(4(20x+20)*5(25x-40))), so:

5x2/(4(20x+20)*5(25x-40))

We multiply all the terms by the denominator


5x2

We add all the numbers together, and all the variables

5x^2

Back to the equation:

+(5x^2)



We calculate terms in parentheses: +(-(-16x2)/(4(20x+20)*5(25x-40))), so:

-(-16x2)/(4(20x+20)*5(25x-40))

We add all the numbers together, and all the variables

-(-16x^2)/(4(20x+20)*5(25x-40))

We multiply all the terms by the denominator


-(-16x^2)

We get rid of parentheses

16x^2

Back to the equation:

+(16x^2)


We add all the numbers together, and all the variables

21x^2-18=0

a = 21; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·21·(-18)
Δ = 1512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{42}}{2*21}=\frac{0-6\sqrt{42}}{42}
=-\frac{6\sqrt{42}}{42}
=-\frac{\sqrt{42}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{42}}{2*21}=\frac{0+6\sqrt{42}}{42}
=\frac{6\sqrt{42}}{42}
=\frac{\sqrt{42}}{7}
$