1/4x^2-5=5

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

1/4x^2-5=5

We move all terms to the left:

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


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

x^2!=0/4

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

1/4x^2-10=0

We multiply all the terms by the denominator


-10*4x^2+1=0

Wy multiply elements

-40x^2+1=0

a = -40; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-40)·1
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-40}=\frac{0-4\sqrt{10}}{-80}
=-\frac{4\sqrt{10}}{-80}
=-\frac{\sqrt{10}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-40}=\frac{0+4\sqrt{10}}{-80}
=\frac{4\sqrt{10}}{-80}
=\frac{\sqrt{10}}{-20}
$