what's (r-4)(r+12)= -16?

Answer 1

r^2+8r-48+16=0
r^2+8r-32=0

r1=-4+4sqrt(3)
r2==-4-4sqrt(3)

Answer 2

You need to get it into normal quadratic form and solve it.

(r-4)(r+12) = -16
r^2 +8r -48 = -16
r^2 +8r -32 = 0

Does it have real roots? Take a look at the determinant b^2 -4ac.

64 - 4*1*-32 = 64 +128 = 192. It's positive so it has real roots

Plug it into the quadratic formula
(-8 +- sqrt(192)) / 2

(-8 - 13.856)/2 = - 10.9282
(-8 + 13.856)/2 = 2.9282

So r = -10.9282 or 2.9282

Answer 3

(r - 4) (r + 12) = - 16

L.H.S. =

r^2 + 12 r - 4r - 48

= r^2 + 8 r - 48

L.H.S. = R.H.S

r^2 + 8 r - 48 = - 16

=> r^2 + 8r -48 + 16 = 0

=> r^2 + 8r - 32 = 0

Solving for 'r' using the formula for roots = - b + or - root of b-4ac whole divided by 2 a

we get r = -8 + 96 ^ (1/2)

or -8 - 96 ^ (1/2)

In other words,

r = -8 + square root of 96

or

-8 - Square root of 96

Answer
= - 8 + 9.9 = 1.9 ( correct to 1 decimal place)
or
= 8 - 9.9 = - 17.9 ( correct to 1 decimal place) ...

Answer = r = 1.9 or minus 17.9

Answer 4

(r-4)(r+12)= -16
r^2 + 8r - 48 - 16 = 0
r^2 + 8r -64 = 0
r^2 + 8r + 16 = 64+ 16
(r +4)^2 = 80
take square roots of both sides
r + 4 =+sqrare root of 80
also r + 4 = - square root of 80
r = - 4+square root of 80
and r = - 4 - square root of 80

Answer 5

r^2 + 8r - 48 = - 16
r^2 + 8r = 32
r^2 + 8r + 16 = 48
r + 4 = ± 4√3
r = - 4 - 4√3, - 4 + 4√3

Answer 6

r² + 8r - 48 = - 16
r² + 8r - 32 = 0
r = [- 8 ± √(64 + 128)] / 2
r = [- 8 ± √(192)] / 2
r = [- 8 ± 8√3] / 2
r = [- 4 ± 4√3 ]
r = 2.93 , r = - 10.93

Answer 7

(r-4)(r+12)=-16
r^2+8r-48+16=0
r^2+8r-32=0