Solve the equation 12y^2+25y+12=0 Answer: y = y=square Write your answers as a list of integers or reduced fractions, with your answers separated by (a) comma(s). For example, if you get 4 and -(2)/(3) as your answers, then enter 4,-2/3 in the box.

To solve the quadratic equation $12y^{2}+25y+12=0$, we can use the quadratic formula:<br /><br />$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 12$, $b = 25$, and $c = 12$.<br /><br />Substituting these values into the quadratic formula, we get:<br /><br />$y = \frac{-25 \pm \sqrt{25^2 - 4(12)(12)}}{2(12)}$<br /><br />Simplifying further, we have:<br /><br />$y = \frac{-25 \pm \sqrt{625 - 576}}{24}$<br /><br />$y = \frac{-25 \pm \sqrt{49}}{24}$<br /><br />$y = \frac{-25 \pm 7}{24}$<br /><br />Therefore, the solutions to the equation $12y^{2}+25y+12=0$ are:<br /><br />$y = \frac{-25 + 7}{24} = \frac{-18}{24} = -\frac{3}{4}$<br /><br />$y = \frac{-25 - 7}{24} = \frac{-32}{24} = -\frac{4}{3}$<br /><br />So, the final answer is: $-\frac{3}{4}, -\frac{4}{3}$.