Question

(2x-1)(3x+1)=0
ans fast please​

Answer 1

Answer:

1/2 or -1/3

Step-by-step explanation:

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

open the bracket

$6x^{2} + 2x - 3x - 1 = 0 \\ 6x ^{2} - x - 1 = 0$

now take the equation

$x = -b +/- \sqrt{b {}^{2} } - 4ac \div 2a$

a=6 b=-1 -b=1 c=-1

$x= 1+/- \sqrt{ ( - 1) {}^{2} } - 4 \times 6 \times - 1 \div 2 \times 6$

$x= 1+/- \sqrt{1 +24 \div 12 }$

$x= 1+/- \sqrt{25 \div 12}$

$x= 1+/-5 \div 2$

$x= 1 + 5 \div 12 = 1/2$

$x = 1 - 5 \div 12 = -1/3$