Question

What are the steps to solve6x^2+13x-5=0

Answer 1

$\begin{gathered} \text{Given} \\ 6x^2+13x-5=0 \end{gathered}$

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.

$\begin{gathered} x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \\ x = \frac{ -13 \pm \sqrt{13^2 - 4(6)(-5)}}{ 2(6) } \\ x=\frac{-13\pm\sqrt{169-(-120)}}{12} \\ x=\frac{-13\pm\sqrt{169+120}}{12} \\ x = \frac{ -13 \pm \sqrt{289}}{ 12 } \\ x = \frac{ -13 \pm 17\, }{ 12 } \end{gathered}$

Solve for the two values of x and we get

$\begin{gathered} x_1=\frac{-13+17\,}{12} \\ x_1=\frac{4}{12} \\ x_1=\frac{1}{3} \\ \\ x_2=\frac{-13-17\,}{12} \\ x_2=\frac{-30}{12} \\ x_2=-\frac{5}{2} \end{gathered}$

Therefore, the solution to the given equation is

$x=\frac{1}{3}\text{ and }x=-\frac{5}{2}$