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

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

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

We move all terms to the left:

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


Domain of the equation: (x-3)!=0

We move all terms containing x to the left, all other terms to the right

x!=3

x∈R


We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


(2x^2-7x+4)=0

We get rid of parentheses

2x^2-7x+4=0

a = 2; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·2·4
Δ = 17
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{-(-7)-\sqrt{17}}{2*2}=\frac{7-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{17}}{2*2}=\frac{7+\sqrt{17}}{4}
$