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

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Answer 1 · 2016-11-13T17:40:47

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#

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

Substitute in the values of a, b, and c:

#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#