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

The roots are $x = - \frac{7}{2}$ and $x = 1$
graph{2x^2+5x-7 [-20, 20, -12,12] [-20, 20, -12, 12]}

#### Explanation:

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
$2 {x}^{2} + 5 x - 7 = 0$
By factoring method,
$2 {x}^{2} + 5 x - 7 = 0$
$\left(2 x + 7\right) \left(x - 1\right) = 0$
by the zero property
$2 x + 7 = 0$ and $x - 1 = 0$
it follows
the roots are
$x = - \frac{7}{2}$ and $x = 1$

It can also be checked from the graph the points $\left(- \frac{7}{2} , 0\right)$ and $\left(1 , 0\right)$
God bless...I hope the explanation is useful.

Mar 2, 2018

Using the quadratic formula, you can find out that the quadratic has two real solutions.

#### Explanation:

By evaluating the discriminant from the quadratic formula (${b}^{2} - 4 a c$), 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:

$\textcolor{w h i t e}{\implies} {b}^{2} - 4 a c$

$\implies {5}^{2} - \left(4 \left(2\right) \left(- 7\right)\right)$

$= 25 - \left(8 \left(- 7\right)\right)$

$= 25 - \left(- 56\right)$

$= 25 + 56$

$= 81$

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.