How do you know how many solutions `2x^2+5x-7=0` has?

One way to find the number of roots is by the graph. It is clear that the graph crosses the x-axis at 2 different values of x. Therefore there are 2 roots.

graph{2x^2+5x-7 [-20, 20, -12,12] [-20, 20, -12, 12]}

The give equation is
`2x^2+5x-7=0`
By factoring method,
`2x^2+5x-7=0`
`(2x+7)(x-1)=0`
by the zero property
`2x+7=0` and `x-1=0`
it follows
the roots are
`x=-7/2` and `x=1`

It can also be checked from the graph the points `(-7/2, 0)` and `(1, 0)`
God bless...I hope the explanation is useful.

By evaluating the discriminant from the quadratic formula (`b^2-4ac`), we can find out if the quadratic has two, one, or no real solutions.

If the discriminant is greater than `0`, that means that the quadratic has `2` real solutions.

Furthermore, if the discriminant is greater than `0` and is a perfect square, the quadratic has two real and rational solutions.

If the discriminant is exactly `0`, then the quadratic has exactly `1` real solution.

Lastly, if the discriminant is less than `0`, then the quadratic does not have any real solutions.

Let's evaluate the discriminant for our quadratic:

`color(white)=>b^2-4ac`

`=>5^2-(4(2)(-7))`

`=25-(8(-7))`

`=25-(-56)`

`=25+56`

`=81`

Since the discriminant is greater than `0`, the quadratic has two real solutions. Also, since it's a perfect square, then two solutions are also rational.