(6x^2-19x-7)/2x-7=4x-3

Solution for (6x^2-19x-7)/2x-7=4x-3 equation:

(6x^2-19x-7)/2x-7=4x-3

We move all terms to the left:

(6x^2-19x-7)/2x-7-(4x-3)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We get rid of parentheses

(6x^2-19x-7)/2x-4x+3-7=0

We multiply all the terms by the denominator


(6x^2-19x-7)-4x*2x+3*2x-7*2x=0

Wy multiply elements

-8x^2+(6x^2-19x-7)+6x-14x=0

We get rid of parentheses

-8x^2+6x^2-19x+6x-14x-7=0

We add all the numbers together, and all the variables

-2x^2-27x-7=0

a = -2; b = -27; c = -7;
Δ = b2-4ac
Δ = -272-4·(-2)·(-7)
Δ = 673
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{-(-27)-\sqrt{673}}{2*-2}=\frac{27-\sqrt{673}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{673}}{2*-2}=\frac{27+\sqrt{673}}{-4}
$