20=(18x-2x^2)/2

Solution for 20=(18x-2x^2)/2 equation:

20=(18x-2x^2)/2

We move all terms to the left:

20-((18x-2x^2)/2)=0

We multiply all the terms by the denominator


-((18x-2x^2)+20*2)=0


We calculate terms in parentheses: -((18x-2x^2)+20*2), so:

(18x-2x^2)+20*2

We add all the numbers together, and all the variables

(18x-2x^2)+40

We get rid of parentheses

-2x^2+18x+40

Back to the equation:

-(-2x^2+18x+40)


We get rid of parentheses

2x^2-18x-40=0

a = 2; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·2·(-40)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{161}}{2*2}=\frac{18-2\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{161}}{2*2}=\frac{18+2\sqrt{161}}{4}
$