Answer :
We have the following equation
[tex]4x^2-12x=91\Rightarrow4x^2-12x-91=0[/tex]We can solve it using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where, in our problem
a = 4
b = -12
c = -91
Therefore
[tex]\begin{gathered} x=\frac{-(-12)\pm\sqrt[]{(-12)^2-4\cdot4\cdot(-91)}}{2\cdot4} \\ \\ x=\frac{12\pm\sqrt[]{144+16\cdot91)}}{8} \\ \\ x=\frac{12\pm\sqrt[]{144+1456}}{8} \\ \\ x=\frac{12\pm\sqrt[]{1600}}{8} \\ \\ x=\frac{12\pm40}{8} \end{gathered}[/tex]Now we have the solutions, we only want the negative one, then
[tex]\begin{gathered} x=\frac{12-40}{8} \\ \\ x=-\frac{28}{8} \\ \\ x=-3.5 \end{gathered}[/tex]The negative solution of the equation is -3.5, then, the correct answer is the letter A.