Let's assume there is at least one solution:
x² + 6x + 10 = 0
Remember the discriminant of a quadratic tells you how many solutions there are. The discriminant is that portion of the quadratic formula under the square root symbol.
d = b² - 4ac
If d > 0, there are 2 real solutions
If d = 0, there is 1 real solution (double root)
If d < 0, there are no real solutions
Indeed if you plug in your values for a, b, c you get a negative discriminant:
d = b² - 4ac
a = 1
b = 6
c = 10
d = 6² - 4(1)(10)
d = 36 - 40
d = -4
Hence there are no real solutions.
Another way you can show this is to rewrite the equation for the parabola in vertex form:
y = (x - h)² + k
y = x² + 6x + 10
To complete the square:
a. Take the coefficient in front of x --> 6
b. Halve it --> 3
c. Square it --> 3² = 9
d. Add it and subtract it:
x² + 6x + 9 - 9 + 10
The first part is a perfect square:
(x² + 6x + 9) + (-9 + 10)
(x + 3)(x + 3) + 1
y = (x + 3)² + 1
In vertex form we know that (-3, 1) is the vertex and since it is upward facing (leading coefficient is positive), it will never go below the x-axis.
Or you could mathematically show that this can never equal zero.
As you can see the square can never be negative. And if you add 1 to that, it will always be positive. It can't ever be zero.
You can perhaps see this more clearly, if you try to set it to zero.
(x + 3)² + 1 = 0
(x + 3)² = -1
A square of a real number can never be negative, so there is no real number solution for x.
A third way would be to graph y = x² + 6x + 10 and show that it never crosses the x-axis, so there is no way it could ever be equal to zero.
Can you think of other ways to prove this? Which way do you prefer?