How do you solve #4x ^ { 2} + 6x - 19= 0#?

1 Answer
Tazwar Sikder
May 17, 2017

#x = frac(- 3 - sqrt(85))(4), frac(- 3 + sqrt(85))(4)#

Explanation:

We have: #4 x^(2) + 6 x - 19 = 0#

Let's solve for #x# using the quadratic formula:

#Rightarrow x = frac(- 6 pm sqrt(6^(2) - 4 (4) (- 19)))(2 (4))#

#Rightarrow x = frac(- 6 pm 2 sqrt(85))(8)#

#Rightarrow x = frac(- 3 pm sqrt(85))(4)#

#therefore x = frac(- 3 - sqrt(85))(4), frac(- 3 + sqrt(85))(4)#

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