1/5(25x-75)+9=1/2(2x+54)

Solution for 1/5(25x-75)+9=1/2(2x+54) equation:

1/5(25x-75)+9=1/2(2x+54)

We move all terms to the left:

1/5(25x-75)+9-(1/2(2x+54))=0


Domain of the equation: 5(25x-75)!=0

x∈R



Domain of the equation: 2(2x+54))!=0

x∈R


We calculate fractions


(2x2/(5(25x-75)*2(2x+54)))+(-5x2/(5(25x-75)*2(2x+54)))+9=0


We calculate terms in parentheses: +(2x2/(5(25x-75)*2(2x+54))), so:

2x2/(5(25x-75)*2(2x+54))

We multiply all the terms by the denominator


2x2

We add all the numbers together, and all the variables

2x^2

Back to the equation:

+(2x^2)



We calculate terms in parentheses: +(-5x2/(5(25x-75)*2(2x+54))), so:

-5x2/(5(25x-75)*2(2x+54))

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 add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+9=0

a = -3; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-3)·9
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-3}=\frac{0-6\sqrt{3}}{-6}
=-\frac{6\sqrt{3}}{-6}
=-\frac{\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-3}=\frac{0+6\sqrt{3}}{-6}
=\frac{6\sqrt{3}}{-6}
=\frac{\sqrt{3}}{-1}
$