40=(0.5x^2)+2x

Solution for 40=(0.5x^2)+2x equation:

40=(0.5x^2)+2x

We move all terms to the left:

40-((0.5x^2)+2x)=0


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

(0.5x^2)+2x

We add all the numbers together, and all the variables

2x+(0.5x^2)

We get rid of parentheses

0.5x^2+2x

Back to the equation:

-(0.5x^2+2x)


We get rid of parentheses

-0.5x^2-2x+40=0

a = -0.5; b = -2; c = +40;
Δ = b2-4ac
Δ = -22-4·(-0.5)·40
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{21}}{2*-0.5}=\frac{2-2\sqrt{21}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{21}}{2*-0.5}=\frac{2+2\sqrt{21}}{-1}
$