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

Question

1715 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-28T11:32:10

#X = 1.387 , -0.720#

#D = b^2 - 4ac#

#-> (-2)^2 -4 xx 3 xx -3#

#D = 40#

Now..

# X = ((-b) +- sqrt D)/(2a)#

#-> X = (-(-2) +- sqrt 40)/(2×3)#

#:. x = (2 +- 6.324)/6#

#-> x = 8.324/6# #or# #x= - 4.324/6#

#:.# #X = 1.387 , -0.720#

Answer 2 · 2017-06-28T11:38:46

Use the #B^2 - 4AC = D#

then #x1 ; x2 = (- B +- sqrt(D))/(2A)#

When #A= 3, B = -2, C = -3#

#D = 4 + 36#

#D = 40#

#x1 = (2+sqrt(40))/6#

#x2 = (2-sqrt(40))/6#