What is the solution set for 2x^2+15=-11x A. - -5,-1.5 B 2.5,3 C. 1.5,5 D -3,-2.5

To solve the quadratic equation \(2x^2 + 15 = -11x\), we first need to rewrite it in standard form \(ax^2 + bx + c = 0\).<br /><br />Starting with the given equation:<br />\[ 2x^2 + 15 = -11x \]<br /><br />Add \(11x\) to both sides to move all terms to one side:<br />\[ 2x^2 + 11x + 15 = 0 \]<br /><br />Now, we can use the quadratic formula to solve for \(x\):<br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />Here, \(a = 2\), \(b = 11\), and \(c = 15\). Plugging these values into the quadratic formula gives:<br />\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 2 \cdot 15}}{2 \cdot 2} \]<br />\[ x = \frac{-11 \pm \sqrt{121 - 120}}{4} \]<br />\[ x = \frac{-11 \pm \sqrt{1}}{4} \]<br />\[ x = \frac{-11 \pm 1}{4} \]<br /><br />This results in two solutions:<br />\[ x = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5 \]<br />\[ x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \]<br /><br />Therefore, the solution set is:<br />\[ \{ -3, -2.5 \} \]<br /><br />So, the correct answer is:<br />D. \(\{ -3, -2.5 \}\)