1/4x^+5/4x+1=0

Solution for 1/4x^+5/4x+1=0 equation:

1/4x^+5/4x+1=0


Domain of the equation: 4x^!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We calculate fractions


4x/16x^2+20x/16x^2+1=0

We multiply all the terms by the denominator


4x+20x+1*16x^2=0

We add all the numbers together, and all the variables

24x+1*16x^2=0

Wy multiply elements

16x^2+24x=0

a = 16; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·16·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*16}=\frac{-48}{32}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*16}=\frac{0}{32}
=0
$