How do you solve #8x^2 + 4x = -5# using the quadratic formula?

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Answer 1 · 2017-04-25T14:47:40

#x=(-1+-3i)/(4)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

where #ax^2+bx+c=0#

Here is your quadratic equation

#8x^2+4x=-5#

Move the #-5# to the other side of the equation to complete your quadratic equation

#8x^2+4x+5=0#

Looking at #ax^2+bx+c=0#
I can see that your values are...

#a=8#
#b=4#
#c=5#

Now just put those values into the quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(4)+-sqrt((4)^2-4(8)(5)))/(2(8))#

#x=(-4+-sqrt(16-160))/(16)#

#x=(-4+-sqrt(-144))/(16)#

Here we notice that we have a negative square root.
This turns into an imaginary number or #i# to remove the negative.

#x=(-4+-isqrt(144))/(16)#

#x=(-4+-12i)/(16)#

Notice we have a common factor of #4# in the numerator and denominator.

#x=(4(-1+-3i))/(4(4))#

Cancel out common factor

#x=(cancel(4)(-1+-3i))/(cancel(4)(4))#

#x=(-1+-3i)/(4)#