# How do you find the discriminant for 2x^2-11x+10=0 and determine the number and type of solutions?

Jul 18, 2017

$\Delta = \sqrt{41}$, so $f$ has two real zeroes.

#### Explanation:

We know that, for any quadratic polynomial of the form $a {x}^{2} + b x + c$, the discriminant, $\Delta$, is given by $\Delta = {b}^{2} - 4 a c$.

If ∆ > 0, then the polynomial will have two real solutions.

If ∆=0, then the polynomial will have one real solution - a double root.

And if ∆ < 0, then the polynomial will have no real roots and two complex roots.

And if ∆ is a perfect square, then the polynomial will have rational roots.

For the given polynomial, ∆=(11^2-4xx2xx10)=41. Since ∆ > 0, but sqrt∆ notin QQ, the polynomial will have two real roots.