Question

Find the values of k for which the equation x2-2(1+3k)x+7(3+2k)=0is7

Answer 1

I have solved it through my way... I don't know it's correct or not..

Answer 2

${x}^{2} - 2(1 + 3k)x + 7(3 + 2k) = 0 \\ {x}^{2} - 2x - 6kx + 21 + 14k = 0 \\ {x}^{2} - 2x - 6kx + 14k = - 21 \\ {7}^{2} - 2(7) - 6k(7) + 14(7) = - 21 \\ 49 - 14 - 42k + 98 = - 21 \\ 35 - 42k + 98 = - 21 \\ - 42k = - 21 - 35 - 98 \\ - 42k = - 154 \\ 42k = 154 \\ k = \frac{154}{42} \\ k = 3.6666........ \\ k = 3.6$

.