Solve the equation: 38x²-9x-5=0?

Factor first:

(19x + 5) * (2x - 1) = 0

Now just read off the roots

x = {-5/19, 1/2}

(19x + 5)(2x - 1) = 0
x = - 5 / 19 , x = 1 / 2

the answer is (19x+5)(2x-1)
then
19x+5=0 2x-1=0
19x=-5 2x=1
x=-5/19 x=1/2

Not even going to try to factor that - I'll go right to the quadratic equation:

x = [ -b ± √(b² - 4ac)] / 2a
x = [ -(-9) ± √((-9)² - 4(38)(-5))] / 2(38)
x = [ 9 ± √(81 + 760)] / 76
x = [ 9 ± √(841)] / 76
x = [ 9 ± 29 ] / 76
x = (9 + 29) / 76, (9 - 29) / 76
x = 38 / 76, -20 / 76
x = 1/2, -5/19

38x^2 - 9x - 5 = 0
(2x - 1)(19x + 5) = 0

2x - 1 = 0
2x = 1
x = 1/2

19x + 5 = 0
19x = -5
x = -5/19

∴ x = 1/2 , x = -5/19

It factors (19x +5 )(2x -1) =0
x = -5/19, 1/2

(19x + 5)(2x - 1) = 0

19x + 5 = 0 or 2x - 1=0

19x=-5 or 2x=1

x = -5/19, x = 1/2

(19x+5)(2x-1)=0
x=-5/19, 1/2

factors of 38: (1, 2, 19, 38)
factors of 5: (1, 5)

(19x + 5) (2x - 1) = 0

19x + 5 = 0 /// and /// 2x-1 = 0

19x+5=0
19x=-5
x= -5/19

2x-1=0
2x=1
x=1/2

x= -5/19 , 1/2