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

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Answer 1 · 2015-07-10T03:32:04

Solve #y = x^2 + 5x + 2 = 0#

#D = d^2 = b^2 - 4ac = 25 - 8 = 16 = 4^2 -> d = +- 4#

#x = -b/(2a) +- d/(2a) = - 5/2 +- 4/2#

Answer 2 · 2015-07-10T03:52:25

#x=(-5+sqrt(17))/2,##(-5-sqrt(17))/2#

#x^2+5x=-2#

Quadratic Equation
Add #2# to both sides of the equation in order to make it fit the form of a quadratic equation: #ax^2+bx+c=0#.

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

#a=1;# #b=5;# #c=2#

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

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

#x=(-5+-sqrt(25-8))/2# =

#x=(-5+-sqrt(17))/2#

#x=(-5+sqrt(17))/2#

#x=(-5-sqrt(17))/2#