((3/2)x)+4=2x-5

Solution for ((3/2)x)+4=2x-5 equation:

((3/2)x)+4=2x-5

We move all terms to the left:

((3/2)x)+4-(2x-5)=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)+4=0

We get rid of parentheses

((+3/2)x)-2x+5+4=0

We multiply all the terms by the denominator


((+3-2x*2)x)+5*2)x)+4*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

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

We get rid of parentheses

-4x^2+20x^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
$