2x(x-3)=32x(x-3)=3

Solution for 2x(x-3)=32x(x-3)=3 equation:

2x(x-3)=32x(x-3)=3

We move all terms to the left:

2x(x-3)-(32x(x-3))=0

We multiply parentheses

2x^2-6x-(32x(x-3))=0


We calculate terms in parentheses: -(32x(x-3)), so:

32x(x-3)

We multiply parentheses

32x^2-96x

Back to the equation:

-(32x^2-96x)


We get rid of parentheses

2x^2-32x^2-6x+96x=0

We add all the numbers together, and all the variables

-30x^2+90x=0

a = -30; b = 90; c = 0;
Δ = b2-4ac
Δ = 902-4·(-30)·0
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-90}{2*-30}=\frac{-180}{-60}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+90}{2*-30}=\frac{0}{-60}
=0
$