3x+14/(x+4)(x-4)=0

Solution for 3x+14/(x+4)(x-4)=0 equation:

3x+14/(x+4)(x-4)=0


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

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

x+4)(x!=4

x∈R


We use the square of the difference formula

x^2+3x-16=0

a = 1; b = 3; c = -16;
Δ = b2-4ac
Δ = 32-4·1·(-16)
Δ = 73
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{-(3)-\sqrt{73}}{2*1}=\frac{-3-\sqrt{73}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*1}=\frac{-3+\sqrt{73}}{2}
$