#y = x^2 + 32x - 4#
To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form #color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c) = 0#, which this equation is.
So we know that:
#color(red)(a = 1)#
#color(magenta)(b = 32)#
#color(blue)(c = -4)#
The quadratic formula is #x = (-color(magenta)(b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a))#.
Now we can plug in the values for #color(red)(a)#, #color(magenta)(b)#, and #color(blue)(c)# into the quadratic formula:
#x = (-color(magenta)(32) +- sqrt((color(magenta)(32))^2 - 4(color(red)(1))(color(blue)(-4))))/(2(color(red)(1)))#
Simplify:
#x = (-32 +- sqrt(1024 + 16))/2#
#x = (-32 +- sqrt(1040))/2#
#x = (-32 +- 4sqrt65)/2#
#x = -16 +- 2sqrt65#
This is the same thing as:
#x = -16 + 2sqrt65# and #x = -16 - 2sqrt65#
because #+-# means "plus or minus."
Hope this helps!
