2x-3=(1/2)(x-3)

Solution for 2x-3=(1/2)(x-3) equation:

2x-3=(1/2)(x-3)

We move all terms to the left:

2x-3-((1/2)(x-3))=0


Domain of the equation: 2)(x-3))!=0

x∈R


We add all the numbers together, and all the variables

2x-((+1/2)(x-3))-3=0

We multiply parentheses ..

-((+x^2+1/2*-3))+2x-3=0

We multiply all the terms by the denominator


-((+x^2+1+2x*2*-3))-3*2*-3))=0


We calculate terms in parentheses: -((+x^2+1+2x*2*-3)), so:

(+x^2+1+2x*2*-3)

We get rid of parentheses

x^2+2x*2*+1-3

We add all the numbers together, and all the variables

x^2+2x*2*-2

Wy multiply elements

x^2+4x^2-2

We add all the numbers together, and all the variables

5x^2-2

Back to the equation:

-(5x^2-2)


We add all the numbers together, and all the variables

-(5x^2-2)=0

We get rid of parentheses

-5x^2+2=0

a = -5; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-5)·2
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*-5}=\frac{0-2\sqrt{10}}{-10}
=-\frac{2\sqrt{10}}{-10}
=-\frac{\sqrt{10}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-5}=\frac{0+2\sqrt{10}}{-10}
=\frac{2\sqrt{10}}{-10}
=\frac{\sqrt{10}}{-5}
$