Given #3b^2-6b+51=0#

We could apply the quadratic formula directly to obtain solutions,

but, noticing that all terms are divisible by #3#, we can reduce the complexity of calculations by replacing the given equation with

#color(white)("XXX")b^2-2b+17=0#

The quadratic formula tells us that for an equation in the form:

#color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0#

the solution(s) are given by:

#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b-4color(red)acolor(green)c))/(2color(red)a)#

The equation we have been given has complicated this situation (probably intentionally) by using #b# instead of #x# as it's variable. This can cause some confusion with the constant #color(blue)b# in the standard form of the quadratic formula.

Let's re-write the quadratic formula with #b# as the variable and different letters as the place-holder variables:

#color(white)("XXX")color(red)pb^2+color(blue)qb+color(green)r=0#

has solutions:

#color(white)("XXX")b=(-color(blue)q+-sqrt(color(blue)q^2-4color(red)pcolor(green)r))/(2color(red)p)#

Using #b^2color(blue)(-2)bcolor(green)(+17)=0# as our equation

we have

#color(white)("XXX")color(red)p=color(red)1# (by default)

#color(white)("XXX")color(blue)q=color(blue)(""(-2))#

#color(white)("XXX")color(green)r=color(green)(17)#

So the solutions are

#color(white)("XXX")b=(color(blue)2+-sqrt((color(blue)(""-2))^2-4 * color(red) 1 * color(green)(17)))/(2 * color(red)1)#

#color(white)("XXX")=(2+-sqrt(-64))/2=(2+-8sqrt(-1))/2 = 1+-4i#