1/2x2+3-4=0

Solution for 1/2x2+3-4=0 equation:

1/2x^2+3-4=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^2-1=0

We multiply all the terms by the denominator


-1*2x^2+1=0

Wy multiply elements

-2x^2+1=0

a = -2; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-2)·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-2}=\frac{0-2\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-4}
=-\frac{\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-2}=\frac{0+2\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-4}
=\frac{\sqrt{2}}{-2}
$