Answer:
C. [tex]x=\frac{7}{3}[/tex] and [tex]x=\frac{-3}{2}[/tex]
Step-by-step explanation:
We have the quadratic equation [tex]6x^{2}-5x-21=0[/tex].
Now, the roots of the quadratic equation [tex]ax^{2}+bx+c=0[/tex] are given by [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex].
So, from the given equation, we have,
a = 6, b = -5, c = -21.
Substituting the values in [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex], we get,
[tex]x=\frac{5\pm \sqrt{(-5)^{2}-4\times 6\times (-21)}}{2\times 6}[/tex]
i.e. [tex]x=\frac{5\pm \sqrt{25+504}}{12}[/tex]
i.e. [tex]x=\frac{5\pm \sqrt{529}}{12}[/tex]
i.e. [tex]x=\frac{5\pm 23}{12}[/tex]
i.e. [tex]x=\frac{5+23}{12}[/tex] and i.e. [tex]x=\frac{5-23}{12}[/tex]
i.e. [tex]x=\frac{28}{12}[/tex] and i.e. [tex]x=\frac{-18}{12}[/tex]
i.e. [tex]x=\frac{7}{3}[/tex] and i.e. [tex]x=\frac{-3}{2}[/tex]
Thus, the value for x are [tex]\frac{7}{3}[/tex] and [tex]\frac{-3}{2}[/tex]
