((4x+1)(x+1))/(x-5)=0

Solution for ((4x+1)(x+1))/(x-5)=0 equation:

((4x+1)(x+1))/(x-5)=0


Domain of the equation: (x-5)!=0

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

x!=5

x∈R


We multiply parentheses ..

((+4x^2+4x+x+1))/(x-5)=0

We multiply all the terms by the denominator


((+4x^2+4x+x+1))=0


We calculate terms in parentheses: +((+4x^2+4x+x+1)), so:

(+4x^2+4x+x+1)

We get rid of parentheses

4x^2+4x+x+1

We add all the numbers together, and all the variables

4x^2+5x+1

Back to the equation:

+(4x^2+5x+1)


We get rid of parentheses

4x^2+5x+1=0

a = 4; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·4·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*4}=\frac{-2}{8}
=-1/4
$