Help pls...

The solutions to the equation

$6x^2 + 10x = 4 - 10x - 6x^2 \text{ can be written in the form }\\ x=\frac{P\pm \sqrt Q}{R}, \text{ where P and R}\\ \text{are relatively prime integers and } R>0. \\ \text{What is the product PQR} ?$

$\begin{array}{|rcll|} \hline 6x^2 + 10x &=& 4 - 10x - 6x^2 \\ 12x^2 +20x -4 &=& 0 \\\\ x &=& \dfrac{-20\pm\sqrt{20^2-4\cdot 12\cdot(-4) } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{400+4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4\cdot 100 +4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4}\sqrt{100+48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{148 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{4\cdot 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm2\sqrt{4}\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm 4\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-4\cdot 5\pm 4\sqrt{37 } } {4\cdot 6} \\ x &=& \dfrac{ 4\cdot (-5\pm \sqrt{37}) } {4\cdot 6} \\ \mathbf{x} &=& \mathbf{\dfrac{ -5\pm \sqrt{37} } {6} } \quad | \quad P=-5,\ Q=37,\ R = 6 \\\\ PQR&=& -5\cdot 37 \cdot 6 \\ \mathbf{PQR} &=& \mathbf{-1110 } \\ \hline \end{array}$

Thank you Alan!

Thank you CPhill!