Respuesta :
In order to solve this quadratic equation, first let's put it in the standard form:
[tex]\begin{gathered} y=ax^2+bx+c \\ \\ -19x=-4x^2+30 \\ 4x^2-19x-30=0 \\ \\ a=4,b=-19,c=-30 \end{gathered}[/tex]Then, let's use the quadratic formula to find the zeros:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{19+\sqrt[]{19^2+480}}{8}=\frac{19+29}{8}=\frac{48}{8}=6 \\ x_2=\frac{19-29}{8}=\frac{-10}{8}=-1.25 \end{gathered}[/tex]So the zeros are x = 6 and x = -1.25, therefore the correct option is A.
