How do you solve #9x^2+11x+18=-10x+8#?

Ohkay, let's do this.
1) Rearrange every term on one side, say to the left.
We now have:
#9x^2 + 11x + 18 +10x -8 = 0#
(the sign-change occurs because the terms "crossed" the equal sign)

Now, you can simplify a bit. We have:
#9x^2 + 21x +10 = 0#

This is a quadratic equation of the form
#ax^2 + bx + c =0# where a,b,c are real numbers (in our case, they are integers, which are also real numbers but enough side-tracking).
To solve this, you need the quadratic formula which is, in its general form:
#x_(1,2) = (-b \pm sqrt(b^2 - 4ac))/(2a)#
It says #x_(1,2)# and there is a #\pm#(plus or minus) sign in this equation because there can be a maximum of two (real) solutions, #x_1# and #x_2# which depend on whether we use the plus sign or the minus sign in that equation.

Anyway, applying this formula to our equation, we need to put #a = 9#, #b=21#, and #c=10#
we get the following:
#x_(1,2) = (-21 \pm sqrt(21^2 - 4*9*10))/(2*9)#
i.e.
#x_(1,2) = (-21 \pm sqrt(441 - 360))/(18)#
i.e.
#x_(1,2) = (-21 \pm sqrt(81))/(18)#
i.e.
#x_(1,2) = (-21 \pm 9)/(18)#
so
#x_1 = -12/18# and #x_2 = -30/18#
i.e.
#x_1 = -2/3# and #x_2 = -5/3#

Q.E.D.