tentukanlah akar akar persamaan kuadrat

[tex]2 {x}^{2} - 12x = - 3 \\ {x}^{2} - 6x = - \frac{3}{2} \\ {(x - 3)}^{2} - 9 = - \frac{3}{2} \\ (x - 3) = \sqrt{9 - \frac{3}{2} } = \sqrt{ \frac{15}{2} } \\ x1 = 3 + \frac{ \sqrt{30} }{2} \\ x2 = 3 - \frac{ \sqrt{30} }{2} [/tex]