Respuesta :

Matching

[tex]\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\quad\text{to}\quad\dfrac{15\pm\sqrt{-215}}{22}[/tex]

We can write several relationships.

... -b = 15

... 2a = 22

... b²-4ac = -215

The first of these gives b=-15; the second gives a=11. Then the third gives

... (-15)² -4·11·c = -215

... -44c = -440

... c = 10

So, the quadratic equation is 11x² -15x +10 = 0.