(3x^2+4x)/2x+9=5

Solution for (3x^2+4x)/2x+9=5 equation:

(3x^2+4x)/2x+9=5

We move all terms to the left:

(3x^2+4x)/2x+9-(5)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

(3x^2+4x)/2x+4=0

We multiply all the terms by the denominator


(3x^2+4x)+4*2x=0

Wy multiply elements

(3x^2+4x)+8x=0

We get rid of parentheses

3x^2+4x+8x=0

We add all the numbers together, and all the variables

3x^2+12x=0

a = 3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*3}=\frac{0}{6}
=0
$