What is the solution set for `2x^2 + 4x +10 = 0`?

We can see that there are no Real solutions by checking the discriminant
`color(white)("XXX")b^2-4ac`

#color(white)("XXX")= 16 - 80 < 0 color(white)("XX")rarrcolor(white)("XX")no Real roots

or
If we look at the graph for the expression, we can see that it does not cross the X-axis and therefore is not equal to zero at any values for `x`:
graph{2x^2+4x+10 [-10, 10, -5, 5]}

For a general form quadratic equation

`color(blue)(ax^2 + bx + c = 0)`

you can determine its roots by using the quadratic formula

`color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))`

Now, you can divide all the terms by `2` to make the calculations easier

`(color(red)(cancel(color(black)(2)))x^2)/color(red)(cancel(color(black)(2))) + (4/2)x + 10/2 = 0`

`x^2 + 2x + 5 = 0`

For this quadratic, you have `a=1`, `b=2`, and `c=5`, which means that the two roots will be

`x_(1,2) = (-1 +- sqrt(2^2 - 4 * 1 * 5))/(2 * 1)`

Notice that the determinant, `Delta`, which is the name given to the expression that's under the square root, , is negative.

`Delta = b^2 - 4ac`

`Delta = 2^2 - 4 * 1 * 5 = -16`

For real numbers, you cannot take the square root of a negative number, which means that the quadratic equation has no real solutions.

Its graph will not intercept the `x`-axis. However, it will have two distinct complex roots.

`x_(1,2) = (-1 +- sqrt(-16))/2`

`x_(1,2) = (-1 +- (i^2 * 16))/2 = (-1 +- i * sqrt(16))/2`

`x_(1,2) = (-1 +- 4i)/2`

The two roots will thus be

`x_1 = (-1 + 4i)/2" "` and `" "x_2 = (-1 - 4i)/2`