How do you solve #2x^2 + 5x + 5 = 0 # using the quadratic formula?

1 Answer
Oct 15, 2015

Answer:

# x = (-5 + 3.87i) / 4, x = (-5 -3.87i) / 4#

Explanation:

Standard Quadratic Equation:
#ax^2 + bx + c = 0#

Where,
a is the constant with x to the power 2.
b is the constant with x to the power 1.
c is the constant.

Standard Quadratic Formula:

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

Given Equation:

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

Here,

a = 2
b = 5
c = 5

# x = (-5 +- sqrt((5)^2 - 4(2)(5))) / (2(2))#
# x = (-5 +- sqrt(25 - 40)) / 4#
# x = (-5 +- sqrt(-15)) / 4#
# x = (-5 +- 3.87i) / 4#
# x = (-5 + 3.87i) / 4, x = (-5 -3.87i) / 4#