How do you solve #x^2+8x=-2# using the quadratic formula?

2 Answers
Apr 16, 2018

Answer:

#x=-4+-sqrt14#

Explanation:

First add #2# to each side so that the equation is in standard form: #y=ax^2+bx+c#.

#x^2+8x+2=0#

The quadratic formula is #(-b+-sqrt(b^2-4ac))/(2a)#.

Here, #a# is #1#, #b# is #8#, and #c# is #2#. Let's plug these numbers into the quadratic formula.

#(-8+-sqrt(8^2-4*1*2))/(2*1)#

#(-8+-sqrt(64-8))/2#

#(-8+-sqrt56)/2#

#(-8+-2sqrt14)/2#

#-4+-sqrt14#

#x=-4+-sqrt14#

Apr 16, 2018

Answer:

#x=-.258, -7.74# Those are averaged btw

Explanation:

first move the two over to the right side
#x^2 +8x +2=0#
Now plug into quadratic formula knowing that a quadratic equation is just #ax^2 +bx +c=0# and the quadratic formula is #x=(-b+-(√b^2-4ac))/(2a)#
#x=(-(8)+-(√(8)^2-4(1)(2)))/(2(1))#
#x=(-8+-(√64-8))/(2)#
#x=(-8+(√56))/2# or #x=(-8-(√56))/2#
Which means #x=-.258, -7.74#