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

Solution for 2x^2-3x+1/2=0 equation:

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

We multiply all the terms by the denominator


2x^2*2-3x*2+1=0

Wy multiply elements

4x^2-6x+1=0

a = 4; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·4·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*4}=\frac{6-2\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*4}=\frac{6+2\sqrt{5}}{8}
$