solve by completing the square?

Question

1) t^-8T+9=0


2)x^+5x-6=0




please show the steps.

Answer 1

1)t^2-8t+9=0
=> t^2-8t= -9
=>(t)^2-2*t*4+(4)^2= -9+16[adding 4^2 or 16 to both sides]
=>(t-4)^2=7
=>t-4=+- sqrt 7[square-rooting both sides]
t= 4+sqrt7 or 4-sqrt7
2)x^2+5x-6=0
=>x^2+5x=6
=>(x)^2+2*x*5/2+(5/2)^2=6+25/4 [adding (5/2)^2 or 25/4 to both sides]
=>(x+5/2)^2=49/4
=>x+5/2=+-7/2 [square rooting both sides]
x=-5/2+7/2 =1
or
-5/2-7/2= -6
therefore x=- 6 or 1 ans

Answer 2

Question 1
t² - 8t + 9 = 0
t = [ 8 ± √(64 - 36) ] / 2
t = [8 ± √(28)] / 2
t = [8 ± 2√(7)] / 2
t = [4 ± √7 ] / 2
Question 2
(x + 6).(x - 1) = 0
x = - 6, x = 1

Answer 3

1.
t^2-8t+9=0
t^2-8t=-9
{t^2-8t+[(1/2)(-8)]^2}=-9+[(1/2)(-8)]^2
{(t-4)^2}=-9+16
(t-4)^2=7
t-4=+/-[7^(1/2)]
t=+/-[7^(1/2)]+4
t=6.646 or t=1.354(up to 3 decimal places)

2.
x^2+5x-6=0
x^2+5x=6
x^2+5x+[(1/2)(5)]^2=6+[(1/2)(5)]^2
(x+5/2)^2=6+25/4
(x+5/2)^2=49/4
x+5/2=+/-(49/4)^(1/2)
x+5/2=+/-(7/2)
x=+/-(7/2)-5/2
x=1 or x=-6