Question

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

Answer 1

This equation has no real solutions.

Explanation:

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.)

Answer 2

Use the quadratic formula to find the complex solutions:

`x = (3+-i sqrt(47))/4`

Explanation:

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`