Use the quadratic formula to solve the quadratic equation. $-$ y 2 + 15 y = 5
A) $\left\{ \frac { 15 \pm \sqrt { 205 } } { 2 } \right\}$
B) $\{ 15 \pm \sqrt { 205 } \}$
C) $\left\{ \frac { 13 \pm \sqrt { 205 } } { 7 } \right\}$
D) $\left\{ \frac { 15 \pm \sqrt { 205 } } { 7 } \right\}$
E) $\left\{ \frac { 18 \pm \sqrt { 205 } } { 2 } \right\}$

Question

Quizplus · Accepted Answer

First, rewrite the equation in standard form: $-y^2 + 15y - 5 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = -1$, $b = 15$, and $c = -5$, we get $y = \frac{-15 \pm \sqrt{225 + 20}}{2(-1)} = \frac{15 \pm \sqrt{245}}{-2}$. Simplifying $\sqrt{245}$ to $\sqrt{205}$ was a mistake; it should remain as $\sqrt{225 + 20} = \sqrt{245}$. Therefore, the correct simplification leads to $y = \frac{15 \pm \sqrt{245}}{-2}$, which is not an option provided, indicating a mistake in my explanation. Correcting this, for $a = -1$, $b = 15$, and $c = -5$, correctly applying the quadratic formula gives $y = \frac{-15 \pm \sqrt{225 - 4(-1)(-5)}}{2(-1)} = \frac{15 \pm \sqrt{225 - 20}}{-2} = \frac{15 \pm \sqrt{205}}{-2}$, which simplifies correctly to the options given. The correct answer, reflecting the proper application of the quadratic formula to the equation $-y^2 + 15y - 5 = 0$, is indeed $y = \frac{15 \pm \sqrt{205}}{2}$, which matches choice A, acknowledging an error in my initial explanation regarding the simplification of the square root term.

Use the Quadratic Formula to Solve the Quadratic Equation −-− Y 2 + 15 Y = 5 A)

Use the Quadratic Formula to Solve the Quadratic Equation $-$ Y 2 + 15 Y = 5
A)