Solving an equation with x and x^2

I need to solve for X in this equation:
9x^2 - x(3W+2L) + (WL/2) = 0

If you could please include steps, that would be much appreciated.
I need to figure out how to turn the equation in question into:
X = L/9 + W/6 - sqrt(4L2 - 6LW + 9W2)/18

Thanks in advance!

Use the quadratic formula.
x = [3W+2L ± √((3W+2L)² - 4·9·WL/2)]/[2·9]
= [3W+2L ± √(9W² - 6WL + 4L²)]/18

9x^2 - x(3W+2L) + (WL/2) = 0
This is a quadratic equation of the form:
ax^2+bx+c
where
a=9
b=-(3W+2L)
c=(WL/2)

so find delta=b^2-4ac
=[-(3W+2L)]^2-4(9)(wl/2)
=(3W+2L)^2-18wl
=9W^2+4l^2+12wl-18wl
=9W^2+4l^2-6wl

=>x1=(-b-√delta)/2a
=>x1=[3w+2l-√(9W^2+4l^2-6wl)]/18
=>x1=w/6+l/9-√(9W^2+4l^2-6wl)/18

and
=>x2=(-b+√delta)/2a
=>x2=w/6+l/9+√(9W^2+4l^2-6wl)/18

Ur answer is half correct, since it has a continuation.
U have found the value of x,yet x has 2 values


If u need more help email at g_zayo@yahoo.com

It's simple!

You know the general form for quadratic formula:

a*x^2 + b*x + c = 0

9*x^2 - x*(3*W + 2*L) + (W*L/2) = 0

a = 9, b = - (3*W + 2*L), c = W*L/2

b^2 - 4*a*c = (3*W + 2*L)^2 - 2*9*W*L = 9*W^2 - 6*W*L + 4*L^2

x1 = ( - ( - (3*W + 2*L)) + sqrt(9*W^2 - 6*W*L + 4*L^2))/(2*9) = (1/6)*W+(1/9)*L+(1/18)*sqrt(9*W^2 - 6*W*L + 4*L^2)

x2 = (1/6)*W + (1/9)*L - (1/18)*sqrt(9*W^2 - 6*W*L + 4*L^2)