How do you find the roots, real and imaginary, of # h=--t^2+16t -4 # using the quadratic formula?

1 Answer
Jun 5, 2017

#color(blue)(h=-8+2sqrt17 or h=-8-2sqrt17#

Explanation:

#h=- -t^2+16t-4#

#h=-(-t^2)+16t-4#

#h=t^2+16t-4#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#color(blue)(a^2+bx+c#

#color(blue)(a=1,b=16,c=-4#

#h=(-(16)+-sqrt((16)^2-4(1)(-4)))/(2(1))#

#(-16+-sqrt(256+16))/2#

#(-16+-sqrt(272))/2#

#(-16+-sqrt(2*2*2*2*17))/2#

#color(blue)(sqrt2 xx sqrt2=2#

#(-16+-4sqrt17)/2#

#(cancel4^color(blue)2(-4+sqrt17))/cancelcolor(blue)2^color(blue)1 or (cancel4^color(blue)2(-4-sqrt17))/cancel2^color(blue)1 #

#2(-4+sqrt17) or 2(-4-sqrt17)) #

#color(blue)(-8+2sqrt17 or -8-2sqrt17#