Factorise 4t^2 + 12t + 80?

divide by 4
t^2+3t+20
it wont factorise so use the quadratic equation
-b± √b^2-4ac
----------------------
2a

a=1 b=3 and c=2-
-3±√(-3)^2 - (4*1*20)
______________________
2

-3±√9-80/2

this has no solution as you cannot get the square root of a negative number

It can't be factorized.

4t^2 + 12t + 80
4(t^2 + 3t + 20)
4(t^2 + 3t + 20)

t=-b+-sqrt(b^2-4ac)/2a
where a=1, b=3, c=20

substituting we get
t=-b+-sqrt(b^2-4ac)/2a
t=-3+-sqrt({3}^2-{4*1*20})/2*1
t=-3+-sqrt(9-{80})/3
t=-3+-sqrt(9-80)/3
t=-3+-sqrt(-71)/3

4t^2 + 12t + 80

divide by 4:
t^2 + 3t + 20 = (t + 1)(t + 3) + 17

at^2+bt+c
t=b+root(b^2-4ac)/2a&t=b-root(b^2-4ac)/2a

4t^2 + 12t + 80
= 4(t^2 + 3t + 20)

4 (t² + 3t + 20)
That`s it pal !

4t^2 + 12t + 80
1) Shorten the polynomial.
4t^2 +12t + 80 / 4
= t^2 + 3t + 20

2) Use the quadratic equation:
-b +- |b^2 - 4 ac| <<<(sq. root) // 2a

3) Substitute.
-(3) + or - /(3)^2 - 4(1)(20)|
divided by 2(1)
= -3 +or- /9 - 80| / 2
= -3 +- /-71| // 2

4) The square root of negative 71 is defined as an imaginary number. The answer cannot be simplified further.

t=
-3 + i/71| /2
or
-3 - i/71| /2