Answer :
[tex]\begin{gathered} \text{Given} \\ 6x^2+13x-5=0 \end{gathered}[/tex]
The given quadratic equation is already in the standard form with the following coefficients.
a = 6, b = 13, and c = -5.
Using these coefficients, solve for x using the quadratic formula.
[tex]\begin{gathered} x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \\ x = \frac{ -13 \pm \sqrt{13^2 - 4(6)(-5)}}{ 2(6) } \\ x=\frac{-13\pm\sqrt{169-(-120)}}{12} \\ x=\frac{-13\pm\sqrt{169+120}}{12} \\ x = \frac{ -13 \pm \sqrt{289}}{ 12 } \\ x = \frac{ -13 \pm 17\, }{ 12 } \end{gathered}[/tex]Solve for the two values of x and we get
[tex]\begin{gathered} x_1=\frac{-13+17\,}{12} \\ x_1=\frac{4}{12} \\ x_1=\frac{1}{3} \\ \\ x_2=\frac{-13-17\,}{12} \\ x_2=\frac{-30}{12} \\ x_2=-\frac{5}{2} \end{gathered}[/tex]Therefore, the solution to the given equation is
[tex]x=\frac{1}{3}\text{ and }x=-\frac{5}{2}[/tex]