How do you use the quadratic formula to solve #16t^2-4t+3=0#?

1 Answer
Sep 24, 2017

Answer:

#t = 1/8+-1/8sqrt(11)i#

Explanation:

Given:

#16t^2-4t+3 = 0#

This is in standard form for a quadratic equation:

#at^2+bt+c = 0#

with #a=16#, #b=-4# and #c=3#

It has roots given by the quadratic formula:

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

#color(white)(t) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(16))(color(blue)(3))))/(2(color(blue)(16)))#

#color(white)(t) = (4+-sqrt(16-192))/32#

#color(white)(t) = (4+-sqrt(-176))/32#

#color(white)(t) = (4+-sqrt(176)i)/32#

#color(white)(t) = (4+-sqrt(16*11)i)/32#

#color(white)(t) = (4+-4sqrt(11)i)/32#

#color(white)(t) = 1/8+-1/8sqrt(11)i#