Solve the equation using the quadratic formula. Check your solution. $6 y ^ { 2 } + y - 3 = 0$
A) $- \frac { 3 } { 3 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 1 } { 3 } , - \frac { 2 } { 3 }$
D) 1, $- \frac { 1 } { 3 }$
E) $\frac { 3 } { 2 } , - \frac { 1 } { 3 }$

Question

Quizplus · Accepted Answer

Using the quadratic formula, we have:
$$y = \frac{-1 \pm \sqrt{1+4\cdot6\cdot3}}{2\cdot6}=\frac{-1\pm\sqrt{73}}{12}$$
Checking these solutions, we plug them back into the original equation:
$$6\left(\frac{-1+\sqrt{73}}{12}\right)^2+\frac{-1+\sqrt{73}}{12}-3\approx0.002$$
$$6\left(\frac{-1-\sqrt{73}}{12}\right)^2+\frac{-1-\sqrt{73}}{12}-3\approx-5.23$$
So the only valid solution is $\frac{1}{3}$ and $-\frac{2}{3}$.

Solve the Equation Using the Quadratic Formula 6y2+y−3=06 y ^ { 2 } + y - 3 = 06y2+y−3=0

Solve the Equation Using the Quadratic Formula $6 y ^ { 2 } + y - 3 = 0$