3x2-3x(2x/2)=8-16

Solution for 3x2-3x(2x/2)=8-16 equation:

3x^2-3x(2x/2)=8-16

We move all terms to the left:

3x^2-3x(2x/2)-(8-16)=0

We add all the numbers together, and all the variables

3x^2-3x(+2x/2)-(-8)=0

We add all the numbers together, and all the variables

3x^2-3x(+2x/2)+8=0

We multiply parentheses

3x^2-6x^2+8=0

We add all the numbers together, and all the variables

-3x^2+8=0

a = -3; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-3)·8
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-3}=\frac{0-4\sqrt{6}}{-6}
=-\frac{4\sqrt{6}}{-6}
=-\frac{2\sqrt{6}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-3}=\frac{0+4\sqrt{6}}{-6}
=\frac{4\sqrt{6}}{-6}
=\frac{2\sqrt{6}}{-3}
$