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

Solution for (3x-8)(x+2)=2x^2 equation:

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

We move all terms to the left:

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

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

We multiply parentheses ..

-2x^2+(+3x^2+6x-8x-16)=0

We get rid of parentheses

-2x^2+3x^2+6x-8x-16=0

We add all the numbers together, and all the variables

x^2-2x-16=0

a = 1; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·1·(-16)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*1}=\frac{2-2\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*1}=\frac{2+2\sqrt{17}}{2}
$