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

Answer 1

Abigail Allen

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

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#

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Answer 2

Julia Adams

To solve the quadratic equation 2x^2 + 5x + 5 = 0 using the quadratic formula, you first identify the coefficients: a = 2, b = 5, and c = 5. Then, apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Substitute the coefficients into the formula and solve for x. The solutions will be the values of x that satisfy the equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.