(12x^2-25x-7)/(3x^2-16x+21)=0

Solution for (12x^2-25x-7)/(3x^2-16x+21)=0 equation:

(12x^2-25x-7)/(3x^2-16x+21)=0


Domain of the equation: (3x^2-16x+21)!=0

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

3x^2-16x!=-21

x∈R


We multiply all the terms by the denominator


(12x^2-25x-7)=0

We get rid of parentheses

12x^2-25x-7=0

a = 12; b = -25; c = -7;
Δ = b2-4ac
Δ = -252-4·12·(-7)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-31}{2*12}=\frac{-6}{24}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+31}{2*12}=\frac{56}{24}
=2+1/3
$