Somebody help me solving this equation!?

Question

I need to solve this equation (3(t-1)^2 )+6 = 2(1-t)

If you can write step by step I will really appreciate that.

Answer 1

3(t-1)² )+6 = 2(1-t)
3(t² - 2t + 1²) + 6 = 2 - 2t
3t² - 6t + 3 + 6 = 2 - 2t
3t² - 6t + 9 - 2 + 2t = 0
3t² -4t +7 = 0
d = -4² - 4.3.7
d = 16 - 84

Answer 2

Let u = t-1. The equation becomes,
3u^2+2u+6 = 0

Apply quadratic formula and t = u+1,
t = (-2±i√68)/6 + 1 = (2 ± i√17)/3

Answer 3

u = t-1

3 u^2 +6 = 2(-u)
3 u^2 + 2 u + 6 = 0

take it from there

Answer 4

nicely, the 4 may well be written as 24/6, that may supply an liquid crystal demonstrate and facilitates you to function it to (x-26)/(6). so which you're left with x/5 = (24 + x - 26)/6. 24 - 26 is -2, so which you have x/5 = (-2 + x)/6. Multiply the two sides via 6 and you have 6x/5 = -2 + x. multiply the two sides via 5 and you have 6x = -10 + 5x (do no longer forget approximately to distribute the 5 to the two the -2 and the x). Now, subtract 5x from the two sides and you're left with x = -10.

Answer 5

3(t^2 - 2t + 1) + 6

3t^2 - 6t + 3 + 6

3t^2 - 6t + 9 = 2 - 2t

3t^2 - 4t + 7 = 0

No real solutions

Use the quad formula and imaginary numbers will result

Answer 6

(3(t-1)^2 )+6 = 2(1-t)

3(t^2 - 2t + 1) + 6 = 2 - 2t {expand (t - 1)^2 on left, distribute on right}

3t^2 - 6t + 3 + 6 = 2 - 2t {distribute on left}

3t^2 - 4t + 7 = 0 {add 2t, subtract 2}

Apply quadratic formula

t = [- b +- sqrt(b^2 - 4ac)]/2a

t = [4 +- sqrt(16 - 84)]/6
t = [4 +- sqrt(- 68)]/6
t = [4 +- 2sqrt(- 17)]/6

t = [2 +- sqrt(17)i]/3

Answer 7

3(t^2-2t+1)+6=2-2t
3t^2-4t+7=0
2t^2+(t-2)^2+3=0
no solutions 2t^2+(t-2)^2+3>=3