faktor persekutuan dari 8x^2-10x-12 dan 16x^2-9

[tex] {8x}^{2} - 10x - 12 = 0 \\ {4x}^{2} - 5x - 6 = 0 \\ \\ x1.2 = \frac{ - ( - 5) \binom{ + }{ - } \sqrt{ {( - 5)}^{2} - 4.4.( - 6) } }{ 2.4} \\ = \frac{5 \binom{ + }{ - } \sqrt{25 +96 } }{8} \\x1 = \frac{5 + \sqrt{121} }{8} \\ x2 = \frac{5 - \sqrt{121} }{8} [/tex]
sedangkan
[tex]1 {6x}^{2} - 9 = 0 \\ (4x {} - 3)(4x + 3) = 0 \\ 4x = 3 \\ x1 = \frac{3}{4} \\ 4 {x} = - 3 \\ x2 = - \frac{3}{4} [/tex]
Semoga membantu..