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.)