1/5(10x+45)-17=-1/4(28x-8)

Solution for 1/5(10x+45)-17=-1/4(28x-8) equation:

1/5(10x+45)-17=-1/4(28x-8)

We move all terms to the left:

1/5(10x+45)-17-(-1/4(28x-8))=0


Domain of the equation: 5(10x+45)!=0

x∈R



Domain of the equation: 4(28x-8))!=0

x∈R


We calculate fractions


(4x2/(5(10x+45)*4(28x-8)))+(-(-5x1)/(5(10x+45)*4(28x-8)))-17=0


We calculate terms in parentheses: +(4x2/(5(10x+45)*4(28x-8))), so:

4x2/(5(10x+45)*4(28x-8))

We multiply all the terms by the denominator


4x2

We add all the numbers together, and all the variables

4x^2

Back to the equation:

+(4x^2)



We calculate terms in parentheses: +(-(-5x1)/(5(10x+45)*4(28x-8))), so:

-(-5x1)/(5(10x+45)*4(28x-8))

We add all the numbers together, and all the variables

-(-5x)/(5(10x+45)*4(28x-8))

We multiply all the terms by the denominator


-(-5x)

We get rid of parentheses

5x

Back to the equation:

+(5x)


a = 4; b = 5; c = -17;
Δ = b2-4ac
Δ = 52-4·4·(-17)
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{33}}{2*4}=\frac{-5-3\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{33}}{2*4}=\frac{-5+3\sqrt{33}}{8}
$