How do you solve #3x^2-10x-8=0#?

You can just eyeball it most of the time, for instance I can quickly see that this will factorise to

#(3x + 2)(x-4) = 0#

In order for the left hand side to be equal to zero then one of the brackets must be equal to zero, so we have

#3x+2 = 0 or x - 4 = 0#

#implies x = -2/3 or 4#

This isn't particularly handy if it's a really nasty equation or if you just can't see the solution so in general you should use the quadratic formula. For quadratic of the form:

#ax^2+bx+c = 0#

#x = (-b+-sqrt(b^2-4ac))/(2a)#

So in this case:

#x = (10+-sqrt(10^2-4(3)(-8)))/(2(3)) = (10+-sqrt(196))/6 = (10+-14)/6#

#therefore x = -4/6 = -2/3 or x = 24/6 = 4#

since you have a number greater than 1 before the #x^2# you will need to multiply that by the #-8# at the end to get true divisible numbers. 3 x -8 = -24, ergo you need to find two numbers that multiply to get -24 and add up to reach -10 (your middle term). in this case, it would be 2 and -12.

align the equation to make further factoring easier:

#3x^2 - 10x - 8 = 0#

#3x^2 - 12x + 2x - 8 = 0# note that #(3x^2 + 2x - 12x - 8 = 0)# isn't as user friendly

factor the #(3x^2 - 12x + 2x - 8 = 0)# into two sets of brakcets:

#(3x (x - 4) + 2(x - 4) = 0)# and group them together:

#(3x + 2)(x - 4) = 0#

now you need to determine what values for #x# will make this equation true.
#3x + 2 = 0#

#3x = -2#

#x = -2/3#

AND

#x - 4 = 0#

#x = 4#