Given:
The quadratic equation ,
[tex]42x^2-29x-5=0[/tex]
The solution of the given quadratic equation can be obtained as,
[tex]\begin{gathered} x=\frac{-(-29)\pm\sqrt[]{(-29)^2-4(42)(-5)}}{2\times42} \\ \text{ =}\frac{29\pm41}{84} \\ x=\frac{29+41}{84}\text{ or x=}\frac{29-41}{84} \\ x=\frac{70}{84}\text{ or x=}\frac{-12}{84} \\ x=\frac{10}{12}\text{ or x=}\frac{-3}{21} \\ x=\frac{5}{6}\text{ or x=}\frac{-1}{7} \end{gathered}[/tex]
Hence, the solutions are,
[tex]x=\frac{5}{6},\text{ }\frac{-1}{7}[/tex]
