How do you solve #3x^2-4x-5=0# using the quadratic formula?

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Answer 1 · 2018-04-05T05:38:12

#x=(2+sqrt19)/3,# #(2-sqrt19)/3#

Solve:

#3x^2-4x-5=0# is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=3#, #b=-4#, #c=-5#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

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

Simplify.

#x=(4+-sqrt76)/6#

Prime factorize #76#.

#x=(4+-sqrt(2^2xx19))/6#

Apply rule: #sqrt(a^2)=a#

#x=(4+-2sqrt19)/6#

Simplify.

#x=(2+-sqrt19)/3#

Solutions for #x#.

#x=(2+sqrt19)/3,# #(2-sqrt19)/3#