(21-2x)x=2

Solution for (21-2x)x=2 equation:

(21-2x)x=2

We move all terms to the left:

(21-2x)x-(2)=0

We add all the numbers together, and all the variables

(-2x+21)x-2=0

We multiply parentheses

-2x^2+21x-2=0

a = -2; b = 21; c = -2;
Δ = b2-4ac
Δ = 212-4·(-2)·(-2)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-5\sqrt{17}}{2*-2}=\frac{-21-5\sqrt{17}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+5\sqrt{17}}{2*-2}=\frac{-21+5\sqrt{17}}{-4}
$