Question

Please find the root of this equation thank you

Answer 1

Since the quadratic formula of ax^2+bx+c is
x=(-b+-(plus OR minus) sqrt (b^2-4ac))/2a, we can get that a is 1 (since 1*x^2 is x^2), b is 4, and c is -16, so x=(-4+-sqrt((-4)^2-4*1*(-16)))/2*1
= (-4+-sqrt(16-(-64)))/2=(-4+-sqrt(80))/2 which is either (-4+sqrt(80))/2 or 
(-4-sqrt(80))/2. Since 80 is divisible by 4 (aka 2 squared), we can write sqrt(80) as 2*sqrt(20) as 4 goes into 80 20 times, getting (-4+2sqrt(20))/2 or
(4-2sqrt(20))/2, and crossing out the 2's we get (-2+sqrt(20)) or (-2-sqrt(20))