(5x^2+24x+20)/(x+4)=0

Solution for (5x^2+24x+20)/(x+4)=0 equation:

(5x^2+24x+20)/(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


(5x^2+24x+20)=0

We get rid of parentheses

5x^2+24x+20=0

a = 5; b = 24; c = +20;
Δ = b2-4ac
Δ = 242-4·5·20
Δ = 176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{11}}{2*5}=\frac{-24-4\sqrt{11}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{11}}{2*5}=\frac{-24+4\sqrt{11}}{10}
$