how many real- number solutions does the equation have?
0=3x^2+18x+27
There are no real-number solutions to the equation 0=3x^2+18x+27.
We can see this by factoring out 3 from the equation:
0 = 3(x^2 + 6x + 9)
Then we can see that the quadratic inside the parentheses is a perfect square:
0 = 3(x + 3)^2
Setting this equal to zero gives us x = -3, but this is the only solution and it is not a real number solution because it makes the denominator equal to zero. Therefore, there are no real-number solutions to the equation.