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:
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 #1x^(2)# for this question.#b# is the number that has the #x# variable associated with it and it would be #-16x#, and #c# is a number by itself and in this case it is 7.
Now, we just plug our values into the equation like this:
#x = (- (-16) +- sqrt((-16)^(2) - 4(1)(7)))/(2(1))#
#x = (16 +-sqrt(256-28))/2#
#x = (16 +- sqrt(228))/2#
For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add 16 and #sqrt(228)# together and divide that by 2:
#x = (16+sqrt(228))/2#
#x = 31.10/2= 15.55#
Now, we subtract #sqrt(228)# from 16 and divide by 2:
#x = (16-sqrt(228))/2#
# x = 0.90/2 = 0.45#
Therefore, the two possible solutions are:
#x = 15.55# and #x = 0.45#
