Question

Please help quick!!

Answer 1

C
6x^2-13X-5=0
(3X+1)(2X-5)
3x-1=0
X=1/3. 2x-5=0. X= 5/2

Answer 2

Answer:

Option A is correct.

$\{\frac{5}{3}, -\frac{1}{3}\}$

Step-by-step explanation:

Given the equation: $6x^2=13x+5$

we can write this as:

$6x^2-13x-5=0$

A quadratic equation is of the form: $ax^2+bx+c=0$

then the solution is given by:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

On comparing with the given equation we have;

a = 6, b = -13 and c = -5 then;

Substitute the given values we have

$x = \frac{-(-13)\pm\sqrt{(-13)^2-4(6)(-5)}}{2(6)}$

$x = \frac{13\pm\sqrt{169+120}}{12}$

Simplify:

$x = \frac{13 \pm\sqrt{289}}{12}$

$x = \frac{13 \pm 17}{12}$

Then;

$x = \frac{13+17}{12}$ and $x = \frac{13-17}{12}$

⇒$x = \frac{20}{12}$ and $x = -\frac{4}{12}$

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

Therefore, the solution for the given equation are:

$\{\frac{5}{3}, -\frac{1}{3}\}$