How do you solve #x² = 5x + 10# using the quadratic formula?

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Answer 1 · 2018-03-16T16:14:03

Read below.

Let's put the variables and the numbers on one side.

We have:

#x^2-5x-10=0# Since this quadratic equation is in the form #ax^2+bx+c=0#, #a=1#, #b=-5# , and #c=-10#

#x=(-b+-sqrt(b^2-4(a)(c)))/(2a)# Plug in the values.

#=>x=(-(-5)+-sqrt((-5)^2-4(1)(-10)))/(2*1)#

#=>x=(5+-sqrt(25+40))/2#

#=>x=(5+-sqrt(65))/2#

Your two answers would be #(5-sqrt(65))/2# and #(5+sqrt(65))/2#