Appendix A - A.4 - Quadratic Equations - Exercises - Page 560: 13

$x = 12, 5$

$3x^{2} - 51x + 180 = 0$ $3(x^{2} - 17x + 60) = 0$ Find two numbers that can be multiplied to get $60$ and can be added together to get $-17$ $= -12$ and $-5$ $3[x^{2} - 12x - 5x + 60] = 0$ $3[x(x-12)-5(x-12)] = 0$ $3(x-12)(x-5)= 0$ $x = 12, 5$