To solve the inequality \(-2x^2 + 12x > 18\), let's follow the necessary steps for working with quadratic inequalities:
1. Rewrite the inequality in standard form:
\(-2x^2 + 12x > 18\)
2. Move all terms to one side to set it to zero:
[tex]\[
-2x^2 + 12x - 18 > 0
\][/tex]
3. Factor the quadratic expression, if possible:
First, recognize that the quadratic expression is
[tex]\[
-2(x^2 - 6x + 9)
\][/tex]
Since, \(x^2 - 6x + 9 = (x - 3)^2\), we can rewrite the inequality as
[tex]\[
-2(x - 3)^2 > 0
\][/tex]
4. Analyze the factored form:
\((x - 3)^2\) is a perfect square and it is always non-negative, meaning \((x - 3)^2 \geq 0\). However, it can never be greater than zero when scaled by a negative coefficient \(-2\), since
[tex]\[
-2(x - 3)^2 \leq 0
\][/tex]
And the expression \(-2(x - 3)^2\) is exactly zero when \(x = 3\), leading to no solutions where \(-2(x - 3)^2 > 0\).
5. Conclusion:
Because \(-2(x - 3)^2\) always evaluates to zero or negative and never strictly positive, there are no values of \(x\) which satisfy \(-2x^2 + 12x > 18\).
Thus, the correct answer is:
[tex]\[
\text{D. no solutions}
\][/tex]
