3x^2-3x(2x/2)=x-16

Solution for 3x^2-3x(2x/2)=x-16 equation:

3x^2-3x(2x/2)=x-16

We move all terms to the left:

3x^2-3x(2x/2)-(x-16)=0

We add all the numbers together, and all the variables

3x^2-3x(+2x/2)-(x-16)=0

We multiply parentheses

3x^2-6x^2-(x-16)=0

We get rid of parentheses

3x^2-6x^2-x+16=0

We add all the numbers together, and all the variables

-3x^2-1x+16=0

a = -3; b = -1; c = +16;
Δ = b2-4ac
Δ = -12-4·(-3)·16
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{193}}{2*-3}=\frac{1-\sqrt{193}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{193}}{2*-3}=\frac{1+\sqrt{193}}{-6}
$