When we encounter a quadratic equation, we often want to write it in the form ax^2+bx+c=0ax2+bx+c=0. In this case this gives 2x^2-3x+7=02x2−3x+7=0.
Sometimes it is easy to spot solutions, for instance is we have x^2+2x+1=0x2+2x+1=0, it is easy to see that (x+1)^2=x^2+2x+1=0(x+1)2=x2+2x+1=0, so it has the solution x=-1x=−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=0ax2+bx+c=0, we can find solutions via
x=(-bpmsqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a.
In this case we get x=(3pmsqrt((-3)^2-4*2*7))/(2*2)=(3pmsqrt(9-56))/4=3/4pm1/4sqrt(-47)x=3±√(−3)2−4⋅2⋅72⋅2=3±√9−564=34±14√−47. Since there is no real number that satisfies this equation, because sqrt(-47) is no real number, the equation 2x^2+7=3x2x2+7=3x has no real solutions. (There are however so called complex solutions, which use a constructed number ii such that i^2=-1i2=−1, but since this may be too abstract I will not go into this.)
