How do you solve #2x^2 + 4x +10 = 0# using the quadratic formula?

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Answer 1 · 2016-05-11T06:00:27

The solutions for the equation are:
#color(blue)( x = -1 + 2i #

# color(blue)(x = -1 - 2i #

#2x^2 + 4x + 10 = 0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=2, b=4 , c=10#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (4)^2-(4 * 2 * 10)#

# = 16 - 80 = - 64 #

The solutions are normally found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-4)+-sqrt(-64))/(2*2) = (-4+-8i)/4#

#= (-4+-8i)/4 = (4 ( -1 +- 2i))/ 4 #

# = (cancel4 ( -1 +- 2i))/cancel 4 #

# = -1 +- 2i #

# x = -1 + 2i #

# x = -1 - 2i #