Respuesta :
Answer:
Option A is correct.
[tex]\{\frac{5}{3}, -\frac{1}{3}\}[/tex]
Step-by-step explanation:
Given the equation: [tex]6x^2=13x+5[/tex]
we can write this as:
[tex]6x^2-13x-5=0[/tex]
A quadratic equation is of the form: [tex]ax^2+bx+c=0[/tex]
then the solution is given by:
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
On comparing with the given equation we have;
a = 6, b = -13 and c = -5 then;
Substitute the given values we have
[tex]x = \frac{-(-13)\pm\sqrt{(-13)^2-4(6)(-5)}}{2(6)}[/tex]
[tex]x = \frac{13\pm\sqrt{169+120}}{12}[/tex]
Simplify:
[tex]x = \frac{13 \pm\sqrt{289}}{12}[/tex]
[tex]x = \frac{13 \pm 17}{12}[/tex]
Then;
[tex]x = \frac{13+17}{12}[/tex] and [tex]x = \frac{13-17}{12}[/tex]
⇒[tex]x = \frac{20}{12}[/tex] and [tex]x = -\frac{4}{12}[/tex]
⇒[tex]x = \frac{5}{3}[/tex] and [tex]x = -\frac{1}{3}[/tex]
Therefore, the solution for the given equation are:
[tex]\{\frac{5}{3}, -\frac{1}{3}\}[/tex]
