1/2x^2+4=20

Solution for 1/2x^2+4=20 equation:

1/2x^2+4=20

We move all terms to the left:

1/2x^2+4-(20)=0


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

x^2!=0/2

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

1/2x^2-16=0

We multiply all the terms by the denominator


-16*2x^2+1=0

Wy multiply elements

-32x^2+1=0

a = -32; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-32)·1
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-32}=\frac{0-8\sqrt{2}}{-64}
=-\frac{8\sqrt{2}}{-64}
=-\frac{\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-32}=\frac{0+8\sqrt{2}}{-64}
=\frac{8\sqrt{2}}{-64}
=\frac{\sqrt{2}}{-8}
$