Tentukan akar akar dari persamaan kuadrat 3x²+7x-40=0 dengan rumus abc​

Jawaban:

3x² + 7x – 40 = 0

-b ± √-b² – 4ac/2a

= -7 ± √ -7² + 4(3)(40)/2(3)

= -7 ± √49 + 480/6

= -7 ± √529/6

= -7 ± 23/6

x1 = -7 + 23/6

= -56/6 + 23/6

= -33/6

= -11/2

x2 = -7 – 23/6

= -56/6 – 23/6

= -79/6