How do you solve #x(x+6)=-15# using the quadratic formula?

1 Answer
Oct 25, 2017

Answer:

No real solutions, x>0

Explanation:

Expand boiii

#x(x+6) #

#x*x# = #x^2#

#x * 6# = #6x#

We have -15, bring that over to make + 15

so the equation is #x^2 + 6x + 15 = 0#

collect terms for the quadratic formula #(-b +- sqrt(b^2 - 4(a)(c)))/(2a)#
a is the number in front of #x^2#
b is the number in front of #x#
c is the number without an #x#, poor lonely number

a = 1
b = 6
c = 15
remember minuses matter, however there are none in this question.

so plug it in
#( -6 +- sqrt(6^2 - 4(1)(15)))/(2(1))#

put into your calculator as is, i'd suggest one from here: https://goo.gl/JtHwiA they are brilliant. (Casio 991ex)

This leaves a negative number in your square root, which means imaginary numbers.. No real solution is obtainable