0.5x+(3/2x-4)=17

Solution for 0.5x+(3/2x-4)=17 equation:

0.5x+(3/2x-4)=17

We move all terms to the left:

0.5x+(3/2x-4)-(17)=0


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

x∈R


We get rid of parentheses

0.5x+3/2x-4-17=0

We multiply all the terms by the denominator


(0.5x)*2x-4*2x-17*2x+3=0

We add all the numbers together, and all the variables

(+0.5x)*2x-4*2x-17*2x+3=0

We multiply parentheses

0x^2-4*2x-17*2x+3=0

Wy multiply elements

0x^2-8x-34x+3=0

We add all the numbers together, and all the variables

x^2-42x+3=0

a = 1; b = -42; c = +3;
Δ = b2-4ac
Δ = -422-4·1·3
Δ = 1752
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:

$\sqrt{\Delta}=\sqrt{1752}=\sqrt{4*438}=\sqrt{4}*\sqrt{438}=2\sqrt{438}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{438}}{2*1}=\frac{42-2\sqrt{438}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{438}}{2*1}=\frac{42+2\sqrt{438}}{2}
$