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

Solution for (2x^2+5x-3)/(x^2-x+2)=0 equation:

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


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

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

x^2-x!=-2

x∈R


We multiply all the terms by the denominator


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

We get rid of parentheses

2x^2+5x-3=0

a = 2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·2·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*2}=\frac{2}{4}
=1/2
$