Use the Quadratic Formula to solve for $y$ . $x ^ { 2 } + 12 x y + 36 y ^ { 2 } - 5 x - y - 3 = 0$
A) $y = \frac { - ( 36 x + 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$
B) $y = \frac { - ( 12 x + 1 ) \pm \sqrt { ( 36 x + 1 ) ^ { 2 } - 144 \left( x ^ { 2 } + 5 x + 3 \right) } } { 72 }$
C) $y = \frac { - ( 36 x - 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x + 3 \right) } } { 72 }$
D) $y = \frac { ( 36 x - 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$
E) $y = \frac { - ( 12 x - 1 ) \pm \sqrt { ( 12 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$

Question

Use the Quadratic Formula to solve for $y$ .​ $x ^ { 2 } + 12 x y + 36 y ^ { 2 } - 5 x - y - 3 = 0$ ​
A) $y = \frac { - ( 36 x + 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$
B) $y = \frac { - ( 12 x + 1 ) \pm \sqrt { ( 36 x + 1 ) ^ { 2 } - 144 \left( x ^ { 2 } + 5 x + 3 \right) } } { 72 }$
C) $y = \frac { - ( 36 x - 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x + 3 \right) } } { 72 }$
D) $y = \frac { ( 36 x - 1 ) \pm \sqrt { ( 36 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$
E) $y = \frac { - ( 12 x - 1 ) \pm \sqrt { ( 12 x - 1 ) ^ { 2 } - 144 \left( x ^ { 2 } - 5 x - 3 \right) } } { 72 }$

Quizplus · Accepted Answer

The given equation is quadratic in $y$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 36$, $b = 12x - 1$, and $c = x^2 - 5x - 3$, we find the correct expression for $y$.

Use the Quadratic Formula to Solve For yyy ​ x2+12xy+36y2−5x−y−3=0x ^ { 2 } + 12 x y + 36 y ^ { 2 } - 5 x - y - 3 = 0x2+12xy+36y2−5x−y−3=0

Use the Quadratic Formula to Solve For $y$ $x ^ { 2 } + 12 x y + 36 y ^ { 2 } - 5 x - y - 3 = 0$