(2x^2-13x+19)/(x-5)=0

Solution for (2x^2-13x+19)/(x-5)=0 equation:

(2x^2-13x+19)/(x-5)=0


Domain of the equation: (x-5)!=0

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

x!=5

x∈R


We multiply all the terms by the denominator


(2x^2-13x+19)=0

We get rid of parentheses

2x^2-13x+19=0

a = 2; b = -13; c = +19;
Δ = b2-4ac
Δ = -132-4·2·19
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{-(-13)-\sqrt{17}}{2*2}=\frac{13-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{17}}{2*2}=\frac{13+\sqrt{17}}{4}
$