1/2x^2-16=41

Solution for 1/2x^2-16=41 equation:

1/2x^2-16=41

We move all terms to the left:

1/2x^2-16-(41)=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^2-57=0

We multiply all the terms by the denominator


-57*2x^2+1=0

Wy multiply elements

-114x^2+1=0

a = -114; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-114)·1
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{114}}{2*-114}=\frac{0-2\sqrt{114}}{-228}
=-\frac{2\sqrt{114}}{-228}
=-\frac{\sqrt{114}}{-114}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{114}}{2*-114}=\frac{0+2\sqrt{114}}{-228}
=\frac{2\sqrt{114}}{-228}
=\frac{\sqrt{114}}{-114}
$