Solve the following equations by using the quadratic formula  ?

The Quadratic Equation is 

x = {-b +/-sqrt[b^2 - 4ac]} / 2a 
Where 
'a'  is the coefficient of 'x^2' ; '1' in this case.
'b' is the coefficient of 'x'  ; '-12' is this case
'c' is the numerical constant '4' in this case. 
Substitute in 
x = { - - 12 +/-sqrt[(-12)^2 - 4(1)(4)]} / 2(1) 
NB  Note the 'double minus' after the curly bracket. 
x = {(+)12 +/- sqrt[144 - 16]} / 2 
x = { 12 +/- sqrt[128]} / 2 
x = { 12 +/- 8sqrt(2)] / 2 
x = 6 +/- 4sqrt(2)
x = 6 + 5.6568.. = 11.6568....
& 
 x = 6 - 5.6568... = 0. 34314......Show more

given

a = 1
b = - 12
c = 4

.........- b ±  √(b^2 - 4ac)
x = ----------------------------
....................2a

.......- (-12) ± √[(-12)^2 - 4(1)(4)]
x = -----------------------------------------
................... 2(1)

.......12 ± √ (128)
x=---------------------
.............2

... 12 ± √(64 * 2)
x=----------------------
.............. 2

.........12 ± 8√(2)
x = --------------------
..............2

x = 6 ± 4√(2)

x = 6 + 4√(2), or 6 - 4√(2)  Answer//...Show more

x^2 - 12x + 4
= (x - 4 sqrt(2) - 6) (x + 4 sqrt(2) - 6)...Show more

x^2 -12x + 4 is NOT an equation!

but, if you want the root values:
x^2 -12x + 4 = 0
==> x = (12 +/- sqrt(144 - 16))/2
==> x = 6 +/- (sqrt(128)/2)
now, 128 = 2^7
so...
==> x = 6 +/- (8sqrt(2)/2)
==> x = 6 + 4sqrt(2) or x = 6 - 4sqrt(2) 
done...Show more