How do you solve #x^2 + 8x + 13 = 0#?

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Answer 1 · 2016-02-25T03:28:05

#x=-4+sqrt 3, -4-sqrt 3#

#x^2+8x+13=0# is a quadratic equation in standard form #ax^2+bx+13#, where #a=1, b=8, c=13#.

The quadratic formula can be used to solve the equation.

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

Substitute the known values into the formula and solve for #x#.

#x=(-8+-sqrt(8^2-(4*1*13)))/(2*1)#

Simplify.

#x=(-8+-sqrt(64-52))/2#

Simplify.

#x=(-8+-sqrt 12)/2#

Factor #sqrt 12#.

#sqrt(2xx2xx3)=#

#sqrt(2^2xx3)=#

#2sqrt 3#

#x=(-8+-2sqrt3)/2#

Simplify.

#x=(cancel(-8)^-4+-cancel(2)^1sqrt 3)/cancel(2)^1#

Simplify.

#x=-4+-sqrt 3#

Solutions for #x#.

#x=-4+sqrt 3#

#x=-4-sqrt 3#