((3/2)x)+1=5-2x

Solution for ((3/2)x)+1=5-2x equation:

((3/2)x)+1=5-2x

We move all terms to the left:

((3/2)x)+1-(5-2x)=0


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

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

((+3/2)x)-(-2x+5)+1=0

We get rid of parentheses

((+3/2)x)+2x-5+1=0

We multiply all the terms by the denominator


((+3+2x*2)x)-5*2)x)+1*2)x)=0


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

(+3+2x*2)x

We add all the numbers together, and all the variables

(2x*2+3)x

We multiply parentheses

4x^2+3x

Back to the equation:

+(4x^2+3x)


Wy multiply elements

-20x^2+(4x^2+3x)=0

We get rid of parentheses

-20x^2+4x^2+3x=0

We add all the numbers together, and all the variables

-16x^2+3x=0

a = -16; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-16)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-16}=\frac{-6}{-32}
=3/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-16}=\frac{0}{-32}
=0
$