1/4m^2-5=0

Solution for 1/4m^2-5=0 equation:

1/4m^2-5=0


Domain of the equation: 4m^2!=0

m^2!=0/4

m^2!=√0

m!=0

m∈R


We multiply all the terms by the denominator


-5*4m^2+1=0

Wy multiply elements

-20m^2+1=0

a = -20; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-20)·1
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-20}=\frac{0-4\sqrt{5}}{-40}
=-\frac{4\sqrt{5}}{-40}
=-\frac{\sqrt{5}}{-10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-20}=\frac{0+4\sqrt{5}}{-40}
=\frac{4\sqrt{5}}{-40}
=\frac{\sqrt{5}}{-10}
$