how do i solve this quadratic equation ?

Question

-12x ^2 + 6x + 6 = 0

Answer 1

-12x ^2 + 6x + 6 = 0
[divide by -6] 2x^2 - x - 1 = 0
(2x + 1)(x - 1) = 0
x = -0.5 or 1

Answer 2

-12x ^2 + 6x + 6 = 0
This is of the form ax^2 + bx + c = 0 here a=-12, b=6 & c=6 and the solution of the QUADRATIC equation is given by
x ={ -b +/- sqrt(b^2 - 4ac) } / 2a
==> x ={ -6 +/- sqrt(36+228) } / -24 = (-6 +/- 18) / -24
==> x = (-6+18) / -24 or x = (-6-18) / -24
==> x = -1/2 or 1

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Answer 3

first factor out the 6. then u get : 6(-2x^2+1x+1). Then use the quadratic formula
(x= -b+ or - the square root of (b^2-4ac)divided by 2a). but this can be factored. so it factors into
6(-2x-1)(x-1)=0. That leaves u with either x=-1/2 or x=1. Hope this cleared things up!

Answer 4

-12x ^2 + 6x + 6 = 0
6(-2x+1+1)=0
6(-2x-1)(x-1)=0
true when -2x-1=0 (x=-1/2)
and when x-1=0 (x=1)
ans x=-1/2, 1

Answer 5

12x² - 6x - 6 = 0
2x² - x - 1 = 0
(2x + 1).(x - 1) = 0
x = - 1/2 , x = 1

Answer 6

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

you need to get your a, b, and c, plug in the numbers and there you go. it's pretty easy.