Simplify the equation.
[tex]\begin{gathered} 2x^2+23=14x \\ 2x^2-14x+23=0 \end{gathered}[/tex]Determine the zeros of the equation by using quadratic formula.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{14\pm\sqrt[\square]{(-14)^2-4\cdot2\cdot23}}{2\cdot2} \\ =\frac{14\pm\sqrt[]{12}}{4} \\ =\frac{2(7\pm\sqrt[]{3})}{4} \\ =\frac{7}{2}\pm\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]THus zeros of the equation is,
[tex]\frac{7}{2}+\frac{\sqrt[]{3}}{2}[/tex]and
[tex]\frac{7}{2}-\frac{\sqrt[]{3}}{2}[/tex]