0=(-0.5x^2)-2x+2

Solution for 0=(-0.5x^2)-2x+2 equation:

0=(-0.5x^2)-2x+2

We move all terms to the left:

0-((-0.5x^2)-2x+2)=0

We add all the numbers together, and all the variables

-((-0.5x^2)-2x+2)=0


We calculate terms in parentheses: -((-0.5x^2)-2x+2), so:

(-0.5x^2)-2x+2

We get rid of parentheses

-0.5x^2-2x+2

Back to the equation:

-(-0.5x^2-2x+2)


We get rid of parentheses

0.5x^2+2x-2=0

a = 0.5; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·0.5·(-2)
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{-(2)-2\sqrt{2}}{2*0.5}=\frac{-2-2\sqrt{2}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*0.5}=\frac{-2+2\sqrt{2}}{1}
$