(x^2+7x+12)/(3x-4)=0

Solution for (x^2+7x+12)/(3x-4)=0 equation:

(x^2+7x+12)/(3x-4)=0


Domain of the equation: (3x-4)!=0

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

3x!=4

x!=4/3

x!=1+1/3

x∈R


We multiply all the terms by the denominator


(x^2+7x+12)=0

We get rid of parentheses

x^2+7x+12=0

a = 1; b = 7; c = +12;
Δ = b2-4ac
Δ = 72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*1}=\frac{-6}{2}
=-3
$