((x^2)-(3x)-(10))/(x+4)=0

Solution for ((x^2)-(3x)-(10))/(x+4)=0 equation:

((x^2)-(3x)-(10))/(x+4)=0


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

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

x!=-4

x∈R


We multiply all the terms by the denominator


(x^2-3x-10)=0

We get rid of parentheses

x^2-3x-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
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{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$