How do you use the discriminant to find the number of real solutions of the following
quadratic equation: $2{x^2} + 4x + 2 = 0?$

Hint:If you have the value of discriminant (${b^2} - 4ac$) of an quadratic equation ($a{x^2} + bx + c = 0$), then you can easily tell about the nature and number of real roots as follows
1. Roots will be real and distinct, if $D > 0$
2. Roots will be real and equal, if $D = 0$
3. Roots will be imaginary or no real roots, if $D < 0$
Where $D$ is the discriminant of the quadratic equation.

Complete step by step solution:
Given quadratic equation $2{x^2} + 4x + 2 = 0$

We have to check how discriminant will give the number of real solutions to this quadratic equation
$2{x^2} + 4x + 2 = 0$ and also we will check our answer too by finding the roots of $2{x^2} + 4x + 2 = 0$

So let us find the discriminant of $2{x^2} + 4x + 2 = 0$,
We know that
$D = {b^2} - 4ac$,

In the above quadratic equation, value of $a,\;b$ and $c$ are $2,\;4$ and $2$ respectively
So putting these values in above expression in order to find discriminant of the given quadratic equation
$
D = {b^2} - 4ac \\
D = {4^2} - 4 \times 2 \times 2 \\
D = 16 - 16 = 0 \\
$

As we know that when the value of $D = 0$ then roots are real and equal. So there are $2$ real and equal roots of the quadratic equation $2{x^2} + 4x + 2 = 0$

Now, let us check if our answer is correct or not,

We know that roots of a quadratic equation are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ where D is discriminant
So putting respective values to find roots,

$
\Rightarrow x = \dfrac{{ - 4 + \sqrt 0 }}{{2 \times 2}}\;\& \; \Rightarrow x = \dfrac{{ - 4 - \sqrt 0 }}{{2
\times 2}} \\
\Rightarrow x = \dfrac{{ - 4}}{4}\;\& \; \Rightarrow x = \dfrac{{ - 4}}{4} \\
\Rightarrow x = - 1\;\& \; \Rightarrow x = - 1 \\
$
$\therefore $ our answer is correct, we got two real and equal roots.

Note: If discriminant comes negative then you got imaginary roots because according to this below equation, the part under the square root will become negative if discriminant is negative. And we know that the negative number under a square root gives imaginary numbers.
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$