# How do you find the discriminant and how many solutions does x^2 + 2x – 2 = 0 have?

May 9, 2015

For an equation in the general form:
$a {x}^{2} + b x + c = 0$
the discriminant is:
$\Delta = {b}^{2} - 4 a c$
and
$\Delta \left\{\begin{matrix}< 0 \rightarrow \text{no Real solutions" \\ =0 rarr "1 Real solution" \\ >0 rarr "2 Real solutions}\end{matrix}\right.$

For ${x}^{2} + 2 x - 2 = 0$
$\Delta = {\left(2\right)}^{2} - 4 \left(1\right) \left(- 2\right) = 12 > 0$
so this equation has 2 Real solutions

May 9, 2015

Your equation is in the form: $a {x}^{2} + b x + c = 0$
Where:
$a = 1$
$b = 2$
$c = - 2$
The discriminant is:
$\Delta = {b}^{2} - 4 a c = 4 - 4 \left(1 \cdot - 2\right) = 4 + 8 = 12 > 0$
Now, if:
1] $\Delta > 0$ you have 2 distinct Real solutions;
2] $\Delta = 0$ you have two Real coincident solutions;
3] $\Delta < 0$ you have no Real solutions.