How do you solve the equation #2x^2+5x-3=0# by using the quadratic formula?

1 Answer
Dec 26, 2016

Answer:

The two solutions are #x = 0.5# and #x = -3#

Explanation:

Since this question is given in standard form, meaning that it follows the form: #ax^(2) + bx + c = 0#, we can use the quadratic formula to solve for x:

http://www.1728.org/quadratc.htm
I think it's worthwhile to mention that #a# is the number that has the #x^2# term associated with it. Thus, it would be #2x^(2)# for this question.#b# is the number that has the #x# variable associated with it and it would be #5x#, and #c# is a number by itself and in this case it is -3.

Now, we just plug our values into the equation like this:

#x = (- (5) +- sqrt((5)^(2) - 4(2)(-3)))/(2(2))#

#x = (-5 +-sqrt(25+24))/4#

#x = (-5 +- sqrt(49))/4#

For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add -5 and #sqrt(49)# together and divide that by 4:

#x = (-5+sqrt(49))/4#
#x = 2/4= 0.5#

Now, we subtract #sqrt(49)# from -5 and divide by 4:

#x = (-5-sqrt(49))/4#
# x = -12/4 = -3#

Therefore, the two possible solutions are:
#x = 0.5# and #x = -3#