Question

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

Answer 1

`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`