Question 8 (1 point) For the equation $$ 2 x^{2}-6 x+c=0 $$, which value of $$ C $$ would produce an equation with no solution?

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1. For a quadratic equation $$ ax^2 + bx + c = 0 $$, the discriminant is given by $$ \Delta = b^2 - 4ac. $$ 2. In the equation $$ 2x^2 - 6x + c = 0 $$, we have $$ a = 2 $$, $$ b = -6 $$, and the constant term is $$ c $$. Thus, the discriminant becomes $$ \Delta = (-6)^2 - 4(2)(c) = 36 - 8c. $$ 3. For the equation to have no real solutions, the discriminant must be negative: $$ 36 - 8c < 0. $$ 4. Solving the inequality: $$ \begin{aligned} 36 - 8c &< 0 \ -8c &< -36 \ c &> \frac{36}{8} \ c &> \frac{9}{2}. \end{aligned} $$ 5. Therefore, any value of $$ c $$ such that $$ c > \frac{9}{2} $$ will produce an equation with no real solutions. Any value of $$ c $$ greater than $$ \frac{9}{2} $$ will make the equation $$ 2x^2 - 6x + c = 0 $$ have no real solutions.

Question 8 (1 point)
For the equation , which value of would produce an
equation with no solution?

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Question 8 (1 point) For the equation , which value of would produce an equation with no solution?