Question

3y²-y=10 find root of quadratic equations​

Answer 1

Answer:--

${3y}^{2} - y = 10 \\ = > {3y}^{2} - y - 10 = 0 \\ = > 3 \times y \times y - y - 10 \\ = > 3 \times y \times y - 6 y+ 5y - 10 \\ = > 3 \times y \times y - (3 \times 2)y + 5y + 5 \times - 2 \\ take \: \: + 3y \: and \: \: 5 \: common \\ 3y(y - 2) \: \:5( y - 2) \\ = > (3y + 5) \: \: (y - 2) \\ = > y = \frac{ - 5}{3} = > y = + 2$