0.5x+1(3/2x-4)=17

Solution for 0.5x+1(3/2x-4)=17 equation:

0.5x+1(3/2x-4)=17

We move all terms to the left:

0.5x+1(3/2x-4)-(17)=0


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

x∈R


We multiply all the terms by the denominator


(0.5x)*2x-17*2x-4)+1(3-4)=0

We add all the numbers together, and all the variables

(+0.5x)*2x-17*2x-4)+1(-1)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

0x^2-17*2x=0

Wy multiply elements

0x^2-34x=0

We add all the numbers together, and all the variables

x^2-34x=0

a = 1; b = -34; c = 0;
Δ = b2-4ac
Δ = -342-4·1·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-34}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+34}{2*1}=\frac{68}{2}
=34
$