Question

Find roots of quadratic equation 12x square-8x+1whole 1 upon3=0mirr

Answer 1

$(12x {}^{2} - 8x + 1) \div 3 = 0 \\ 12 x {}^{2} - 8x + 1 = 0 \\ 12x {}^{2} - 6x - 2x + 1 = 0 \\ 6x(2x - 1) - 1(2x - 1) = 0 \\ (2x - 1)(6x - 1) = 0 \\ x = 1 \div 2 \\ x = 1 \div 6$