1/2x^2-5=17

Solution for 1/2x^2-5=17 equation:

1/2x^2-5=17

We move all terms to the left:

1/2x^2-5-(17)=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^2-22=0

We multiply all the terms by the denominator


-22*2x^2+1=0

Wy multiply elements

-44x^2+1=0

a = -44; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-44)·1
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{11}}{2*-44}=\frac{0-4\sqrt{11}}{-88}
=-\frac{4\sqrt{11}}{-88}
=-\frac{\sqrt{11}}{-22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{11}}{2*-44}=\frac{0+4\sqrt{11}}{-88}
=\frac{4\sqrt{11}}{-88}
=\frac{\sqrt{11}}{-22}
$