How do you solve #2x^2-x-4=0# using the quadratic formula?

Question

Eli Assouline · Accepted Answer

The solutions are:
#color(blue)(x=(1+sqrt(33))/4 , x=(1-sqrt(33))/4# The equation #2x^2−x−4# : is of the form #color(blue)(ax^2+bx+c=0# where: #a=2, b=-1, c=-4# The Discriminant is given by: #Delta=b^2-4*a*c# # = (-1)^2 - (4)*(2)(-4) # #=1+32# #=33# As #Delta>0# there are two solutions, The solutions are found using the formula: #x=(-b+-sqrtDelta)/(2*a)# #x = (-(-1)+-sqrt(33))/(2*2) = (1+-sqrt(33))/4# The solutions are: #color(blue)(x=(1+sqrt(33))/4 , x=(1-sqrt(33))/4#

Ethan Adkins · Answer

undefined To solve the quadratic equation $2x^2 - x - 4 = 0$ using the quadratic formula, you would first identify the coefficients $a$, $b$, and $c$, which are $2$, $-1$, and $-4$ respectively. Then you would substitute these values into the quadratic formula:

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

Substitute $a = 2$, $b = -1$, and $c = -4$ into the formula:

$$x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(2)(-4)}}}}{{2(2)}}$$

$$x = \frac{{1 \pm \sqrt{{1 + 32}}}}{{4}}$$

$$x = \frac{{1 \pm \sqrt{33}}}{{4}}$$

So the solutions to the quadratic equation $2x^2 - x - 4 = 0$ using the quadratic formula are:

$$x = \frac{{1 + \sqrt{33}}}{{4}} \quad \text{and} \quad x = \frac{{1 - \sqrt{33}}}{{4}}$$