(4x2-3x+5)/(x2-2x+13)=2

Solution for (4x2-3x+5)/(x2-2x+13)=2 equation:

(4x^2-3x+5)/(x2-2x+13)=2

We move all terms to the left:

(4x^2-3x+5)/(x2-2x+13)-(2)=0


Domain of the equation: (x2-2x+13)!=0

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

x2-2x!=-13

x∈R


We add all the numbers together, and all the variables

(4x^2-3x+5)/(+x^2-2x+13)-2=0

We multiply all the terms by the denominator


-2*(+x^2-2x+13)+(4x^2-3x+5)=0

We multiply parentheses

-2x^2+4x+(4x^2-3x+5)-26=0

We get rid of parentheses

-2x^2+4x^2+4x-3x+5-26=0

We add all the numbers together, and all the variables

2x^2+x-21=0

a = 2; b = 1; c = -21;
Δ = b2-4ac
Δ = 12-4·2·(-21)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*2}=\frac{-14}{4}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*2}=\frac{12}{4}
=3
$