Use the quadratic formula to solve the equation?

Question

3x^2 - 16x = 3

Answer 1

3x^2 - 16x - 3 = 0

a = 3
b = -16
c = -3

x = [-b +- sqrt (b^2 - 4ac)] / 2a
= [-(-16) +- sqrt ((-16)^2 - 4(3)(-3))] / 2(3)
= [16 +- sqrt (256 + 36)] / 6
= [16 +- sqrt 292] / 6
= [16 +- sqrt (4*73)] / 6
= [16 +- 2 sqrt 73] / 6
= [8 +- sqrt 73] / 3

Answer 2

The quadratic formula is x=[-b+/- sqrt (b^2 - 4ac)]/2a.
3x^2-16x=3 is the same as 3x^2-16x-3=0.
So, a=3, b=-16, and c=-3.
Thus, x=[--16+/- sqrt ((-16)^2 - 4*3*-3)]/2*3.
x=[16+/- sqrt (256 + 36)]/6.
x=[16+/- sqrt (292)]/6.
x=[8+/- sqrt (73)]/3.
Thus, the 2 solutions are
x =(8 + sqrt 73) / 3 or
x =(8 - sqrt 73) / 3.

Answer 3

First, make it equals 0 on 1 side.
3x^2 -16x -3 =0
Then apply formula accordingly, take a=3, b=-16 and c=-3
(There is a plus or minus in the formula, do it separately to get 2 results.)

Answer 4

3x^2-16x-3=0
16+-radical 16^2-4(3)(-3)/2(3)
= 16+-radical 256+36/6
=16+-2radical 73/6
=(8+-radical73/3)plug into your calculator get the answers

Answer 5

3x² - 16x - 3 = 0
x = [16 ±√( (-16)² + 36) ] / 6
x = [16 ± √292 ] / 6
x = [ 16 ± 2√73 ] / 6
x = [ 8 ± √73 ] / 3
x = 5.51 , x = - 0.18

Answer 6

Step one know the formula:

ax^2 + bx +c =0

-b +- sqrt(b^2 - 4ac)
----------------------------
2a

Step 2: get everything on one side of the equation.

3x^2-16x -3 =0

Step 3: plug in
we see that a=3 b=-16 and c=-3

-(-16) +- sqrt((-16)^2 - 4(3)(-3))
---------------------------------------------
2(-3)

=

16 +- sqrt(256 + 36)
------------------------------
-6

=

16 +- sqrt(292)
-----------------------
-6

=

16+-2sqrt(73)
-------------------
-6

=

2(8 +- sqrt(73))
---------------------
-6

=
8 +- sqrt(73)
--------------------
-3

so
x=
8 - sqrt(73)
--------------------
-3

and x=
8 + sqrt(73)
--------------------
-3

Answer 7

3x^2-16x=3
<=>3x^2-16x-3=0
a=3
b=-16
c=-3
Delta=b^2-4ac
=(-16)^2-4(3)(-3)
=292
Square root of Delta=Square root of 292=2 square root of 73
=>there will be 2 answer ('cause the square root of Delta>0)
x'=(-b-square root of Delta)/2a
x'={-(-16)-(2 square root of 73)}/2(3)
x'=(16- 2 square root of 73)/6
x'={2(8- square root of 73)}/6
x'=8- square root of 73/3

x''=-b+(square root of Delta)/2a
x''={-(-16)+2 square root of 73}/2(3)
x''=(16+2 square root of 73)/6
x''={2(8+square root of 73)}/6
x''=8+square root of 73/3
The answers are:
x=(8- square root of 73)/3
or
x=(8+ square root of 73)/3

Answer 8

3x^2-16x-3=o a=3 b=-16 c=-3

we use delta'=(-b/2)^2-ac=73

x=(-b/2+-radical73)/a=(8+-radical73)/3