2x(x-3)=9x^2-8x

Solution for 2x(x-3)=9x^2-8x equation:

2x(x-3)=9x^2-8x

We move all terms to the left:

2x(x-3)-(9x^2-8x)=0

We multiply parentheses

2x^2-6x-(9x^2-8x)=0

We get rid of parentheses

2x^2-9x^2-6x+8x=0

We add all the numbers together, and all the variables

-7x^2+2x=0

a = -7; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-7)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-7}=\frac{-4}{-14}
=2/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-7}=\frac{0}{-14}
=0
$