1/2x^2-42=8

Solution for 1/2x^2-42=8 equation:

1/2x^2-42=8

We move all terms to the left:

1/2x^2-42-(8)=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^2-50=0

We multiply all the terms by the denominator


-50*2x^2+1=0

Wy multiply elements

-100x^2+1=0

a = -100; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-100)·1
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-100}=\frac{-20}{-200}
=1/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-100}=\frac{20}{-200}
=-1/10
$