Quadratic equations questions, help please!?

X^2=7X

X^2 - 7X = 0

take X out as it is common,
X(X-7) = 0
so,
X=0, X-7= 0

the answer is X=0,X=7


X = 2 / (6X+1)

X(6X+1)=2
6X^2 + X =2
6X^2 + X - 2 =0

6X^2 + 4x -3X -2 =0

2X(3X+2) - (3X+2) = 0

(3X+2) (2X-1) = 0

(3X+2) =0 (2X-1) =0
the answer is:

X = -2/3 X = 1/2

x² = 7x
x² − 7x = 0
x(x − 7) = 0
x = 0 or x − 7 = 0 i.e. x = 7

I can't understand the second question.

x^2=7x
divide by x x=7

x= 2/6x+1
x*(6x+1)=2
6x^2+x-2=0

this is a quadratic equations determinant is 1+48=49

roots x1 =(-7 -1)/12=-2/3 x2 = (-1+7)/2=1/2

it works

x^2=7x
-----------
x1,2 = (7 ± Sqrt( 49 ) / 2

x1 = (7 + 7) / 2 = 7
x2 = (7 - 7) / 2 = 0


x = 2/(6x + 1)
------------------
6x^2 + x - 2 = 0

x1,2 = ( -1 ± Sqr( 1 + 48 ) ) / 12
x1 = (-1 + 7) / 12 = 1/2
x2 = (-1 - 7) / 12 = - 2/3

0, 7

Question 1

x ² - 7x = 0

x (x - 7) = 0

x = 0 , x = 7

Question 2

x = 2 / (6x + 1)

6 x ² + x = 2

6 x ² - x - 2 = 0

(3x - 2)(2x + 1) = 0

x = 2/3 , x = - 1/2

x^2-7x=0
x=0 & x=7

1)
x^2 = 7x
x^2 - 7x = 0
x(x - 7) = 0

x = 0

x - 7 = 0
x = 7

∴ x = 0 , 7

= = = = = = = =

2)
x = 2/(6x + 1)
x(6x + 1) = 2
6x^2 + x - 2 = 0
6x^2 + 4x - 3x - 2 = 0
(6x^2 + 4x) - (3x + 2) = 0
2x(3x + 2) - 1(3x + 2) = 0
(3x + 2)(2x - 1) = 0

3x + 2 = 0
3x = -2
x = -2/3

2x - 1 = 0
2x = 1
x = 1/2 (0.5)

∴ x = -2/3 , 1/2 (0.5)