((4x^2-5)/3x)=2x+7

Solution for ((4x^2-5)/3x)=2x+7 equation:

((4x^2-5)/3x)=2x+7

We move all terms to the left:

((4x^2-5)/3x)-(2x+7)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

((4x^2-5)/3x)-2x-7=0

We multiply all the terms by the denominator


((4x^2-5)-2x*3x)-7*3x)=0


We calculate terms in parentheses: +((4x^2-5)-2x*3x), so:

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

Wy multiply elements

-6x^2+(4x^2-5)

We get rid of parentheses

-6x^2+4x^2-5

We add all the numbers together, and all the variables

-2x^2-5

Back to the equation:

+(-2x^2-5)


Wy multiply elements

(-2x^2-5)-21x=0

We get rid of parentheses

-2x^2-21x-5=0

a = -2; b = -21; c = -5;
Δ = b2-4ac
Δ = -212-4·(-2)·(-5)
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{401}}{2*-2}=\frac{21-\sqrt{401}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{401}}{2*-2}=\frac{21+\sqrt{401}}{-4}
$