How do you solve #2x^2+3x+5=0# using the quadratic formula?

Answer 1

Christian Antunez

Please see the explanation for steps leading to:

#x = -3/4 + (sqrt(31)i)/4# and #x = -3/4 - (sqrt(31)i)/4#

Comparing with the standard form, #ax^2 + bx + c#, we see that:

#a = 2# #b = 3# #c = 5#

The formula for quadratic equations is

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

Enter the values of a, b, and c as substitutes:

#x = {-3 +-sqrt(3^2 - 4(2)(5))}/(2(2))#

Simplify:

#x = {-3 +-sqrt(-31)}/(4)#

We must factor out #sqrt(-1)# and write it as #i#

#x = -3/4 + (sqrt(31)i)/4# and #x = -3/4 - (sqrt(31)i)/4#

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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.

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