solve x^2-2x+12=0 using the quadratic formula.....?

Question

i have no idea how to do this...
no me gusta math!!

=]

Answer 1

the formula is -b +or- the square root of b^2-4ac all over 2a
a=1
b=-2
c=12
(2+,-√4-48)/2
the answer would be (1+,-i√11)/2

Answer 2

Quadratic equations are of the general form:

ax^2+bx+c=0 where a<>0.

In your case x^2-2x+12=0 you have:
a=1
b=-2
c=12

We culculate:

D=b^2-4ac=(-2)^2-4*1*12=4-48=-44<0

So, your equation has not any real roots, but the complex:
x1=(2+i*sqrt(44))/2 = 1 +i*sqrt(11) and
x2=(2-i*sqrt(44))/2 = 1 -i*sqrt(11)

Answer 3

the quadratic formula is neg B plus or minus the square root of B squared minus 4AC all over 2 A...
so you need the plug them all in
-(-2)+/- sqr root((-2)^2-4(1)(12)) all divided by 2(1)
do whats in the sqr root first so you get 4-48=-44
since what you get is negative it means you get an imaginary number (i), which means it doesnt go through the x axis when graphed
now to solve it, you do the rest of the problem which is (2+/- sqr root(-44)) / 2
the sqr root of -44 is simplified to 2i((sqr rt)-11)
that divided by 2 is i(sq rt -11) and 2divided by 2 is one so you end up with x=1+/-i(sq rt-11)
sorry that its so confusing =(

Answer 4

x² - 2x+12=0

ax² + bx + c = 0
x = (-b +- √(b² - 4ac))/2a

So in your equation
x = (2 +- √(4 - 48))/2
x = (2 +- √(-44))/2

Take 4 out of the radical and it's root is 2

x = (2 +- 2√(-11))/2

Take the root of -1 outside of the radical, it is i

x = (2 +- 2i√(11))/2

Divide numerator and denominator by 2

x = 1 + i√(11), x = 1 - i√(11)
.

Answer 5

-b±√b² - 4ac 2±√-2² - 4(1)(12)
----------------- -----------------------
2a 2(1)

^ ^ ^
Quadratic your equation plugged in
formula


the x² term = a
the x term = b
the # term = c

Just plug it in then use order of operations to simplify

by the way the ( ) means to multiply