How do you solve #2x^2 + 7 = 3x#?

When we encounter a quadratic equation, we often want to write it in the form #ax^2+bx+c=0#. In this case this gives #2x^2-3x+7=0#.

Sometimes it is easy to spot solutions, for instance is we have #x^2+2x+1=0#, it is easy to see that #(x+1)^2=x^2+2x+1=0#, so it has the solution #x=-1#, however, in this case, this is not easy to see, so we need something called the quadratic formula.

When we again look at the general form #ax^2+bx+c=0#, we can find solutions via
#x=(-bpmsqrt(b^2-4ac))/(2a)#.
In this case we get #x=(3pmsqrt((-3)^2-4*2*7))/(2*2)=(3pmsqrt(9-56))/4=3/4pm1/4sqrt(-47)#. Since there is no real number that satisfies this equation, because sqrt(-47) is no real number, the equation #2x^2+7=3x# has no real solutions. (There are however so called complex solutions, which use a constructed number #i# such that #i^2=-1#, but since this may be too abstract I will not go into this.)

First subtract #3x# from both sides to get:

#2x^2-3x+7 = 0#

This is in the form #ax^2+bx+c = 0#, with #a=2#, #b=-3# and #c=7#.

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = (-3)^2 - (4xx2xx7) = 9 - 56 = -47#

Since #Delta < 0# the quadratic has no solutions in real numbers. It has a pair of distinct complex solutions which are complex conjugates of one another.

#x = (-b+-sqrt(Delta))/(2a) = (3+-i sqrt(47))/4#