How do you solve y^2 - 4y - 21 = 0 using the quadratic formula?

1 Answer

x_1=7
x_2=-3

Explanation:

So basically the quadratic equation is x=(-b+-sqrt(b^2-4ac))/(2a)
Here x=y, a=1, b=-4, c=-21
Plug that in and we get:
x=(-(-4)+-sqrt(4^2-4*1*(-21)))/(2*1)

x=(4+-sqrt(16+84))/2

x=(4+-sqrt100)/2

x=(4+-10)/2

x=2+-5

There are two possible values for x.
x_1=7
x_2=-3