1/4x^2+19=300

Solution for 1/4x^2+19=300 equation:

1/4x^2+19=300

We move all terms to the left:

1/4x^2+19-(300)=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^2-281=0

We multiply all the terms by the denominator


-281*4x^2+1=0

Wy multiply elements

-1124x^2+1=0

a = -1124; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1124)·1
Δ = 4496
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:

$\sqrt{\Delta}=\sqrt{4496}=\sqrt{16*281}=\sqrt{16}*\sqrt{281}=4\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{281}}{2*-1124}=\frac{0-4\sqrt{281}}{-2248}
=-\frac{4\sqrt{281}}{-2248}
=-\frac{\sqrt{281}}{-562}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{281}}{2*-1124}=\frac{0+4\sqrt{281}}{-2248}
=\frac{4\sqrt{281}}{-2248}
=\frac{\sqrt{281}}{-562}
$