How do you solve #x(x+6)=-2# using the quadratic formula?

Question

Nathan Axel · Accepted Answer

# x= -0.354 #
OR
# x = -5.645# #x(x+6)=-2# Putting it in formal format: #x^2 +6x+2=0# Compare with equation #ax^2 +bx+c=0# #a = 1# #b=6# #c=2# Quadratic formula: #x=( -b +-sqrt(b^2 -4ac))/(2a)# #x=-0.354248688~~-0.354#---truncated and approximate value. Or #x =-5.645751311~~-5.645#---truncated and approximate value. #therefore x= -0.354 # OR # x = -5.645#

Nathan Axel · Answer

#x = -3 + root2 7 or approx -0.3542... #
#x = -3 - root2 7 or approx -5.6457... # Expand and rearnage; #x^2 + 6x = -2 # #x^2 + 6x + 2 = 0# Then apply to the quadratic formula; # a =1, b = 6, c=2# #x = (-6+root2 ( 6^2 - ( 4*1*2) ))/2# or #x = (-6-root2 ( 6^2 - ( 4*1*2) ))/2# To give #x = (-6+root2 ( 28 ))/2# or #x = (-6-root2 ( 28 ))/2# Hence #x = (-6+2root2 ( 7 ))/2# or #x = (-6-2root2 ( 7 ))/2# Finally giving; #x = -3 + root2 7 or approx -0.3542... # #x = -3 - root2 7 or approx -5.6457... #

Jace Anderson · Answer

undefined To solve the equation $x(x+6)=-2$ using the quadratic formula, first, express the equation in the standard form $ax^2 + bx + c = 0$. In this case, $a = 1$, $b = 6$, and $c = 2$. Then, substitute these values into the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$. After substituting the values, solve for $x$ by performing the arithmetic operations indicated.