91Ó°ÊÓ

If $a{x}^2+bx+c=0$, with $a≠0$, then $x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.

It is given that $6x^2=5x+6$which can also be written as $6x^2−5x−6=0$. Using quadratic formula $x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$, substitute 6 for a, $-5$ for b and $-6$ for c.

Put the values in the formula and get $x=\frac{−\left(−5\right)Â\pm \sqrt{{\left(−5\right)}^2−4×\left(6\right)×\left(−6\right)}}{2\left(6\right)}$.

Simplify the expression for x.

$$\begin{gathered} x=\frac{−\left(−5\right)Â\pm \sqrt{{\left(−5\right)}^2−4×\left(6\right)×\left(−6\right)}}{2\left(6\right)} \\ =\frac{5Â\pm \sqrt{25+144}}{12} \\ =\frac{5Â\pm \sqrt{169}}{12} \\ =\frac{5Â\pm 13}{12} \end{gathered}$$

Therefore, either $x=\frac{5+13}{12}=\frac{18}{12}$or $x=\frac{5−13}{12}=\frac{−8}{12}$that is., either $x=\frac{3}{2}$or $x=−\frac{2}{3}$. Here, the $x$is having two values $\frac{3}{2}$and $−\frac{2}{3}$. Since, $x$is a length of a segment that could not be negative or zero, therefore, the value of $x$is $\frac{3}{2}$.