How do you solve #3x^2 - 2x = 4# using the quadratic formula?

Question

Nathan Axel · Accepted Answer

The solutions are :
#color(blue)(x=(1+sqrt13)/3#

#color(blue)(x=(1-sqrt13)/3# #3x^2-2x-4=0# The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=3, b=-2, c=-4# The Discriminant is given by: #Delta=b^2-4*a*c# # = (-2)^2-(4*3*-4)# # = 4 +48 = 52# The solutions are found using the formula #x=(-b+-sqrtDelta)/(2*a)# #x = (-(-2)+-sqrt(52))/(2*3) = (2+-sqrt(52))/6# Upon further simplification #sqrt52= sqrt(2*2*13)= 2sqrt13# So, #x = (2+-2sqrt(13))/6 = (cancel2(1+-sqrt13))/cancel6# #= (1+-sqrt13)/3# The solutions are : #color(blue)(x=(1+sqrt13)/3# #color(blue)(x=(1-sqrt13)/3#

Hazel Anglin · Answer

undefined The quadratic formula is used to solve equations of the form ax^2 + bx + c = 0. For the given equation 3x^2 - 2x = 4, rewrite it in the form ax^2 + bx + c = 0:

$$3x^2 - 2x - 4 = 0$$

Here, $a = 3$, $b = -2$, and $c = -4$. Now, apply the quadratic formula:

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

Plug in the values of $a$, $b$, and $c$:

$$x = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4 \cdot 3 \cdot (-4)}}}}{{2 \cdot 3}}$$

$$x = \frac{{2 \pm \sqrt{{4 + 48}}}}{{6}}$$

$$x = \frac{{2 \pm \sqrt{{52}}}}{{6}}$$

So, the solutions are:

$$x = \frac{{2 + \sqrt{{52}}}}{{6}}$$

$$x = \frac{{2 - \sqrt{{52}}}}{{6}}$$

Simplify the radical if possible.