Solutions for a quadratic equation

I need to solve
x^2 + 6x = 16 and i completely forgot how. Can someone show me the steps?

Lots of different methods....

By completing the square:

x^2 + 6x = 16

1(x^2 + 6x) = 16

1(x^2 + 6x + 3^2 - 3^2) = 16

1(x^2 + 6x + 9 -9) = 16

1(x^2 + 6x + 9) + 1(-9) = 16

1(x^2 + 6x + 9) - 9 = 16

(x+3)(x+3) = 16 + 9

(x+3)^2 = 25

√(x+3)^2 = +/- √25

x + 3 = +/- 5

x = -3 +/- 5

x1 = -3 - 5

x1 = -8

x2 = -3 + 5

x2 = 2

THEREFORE x = -8, 2






By quadratic formula:

x^2 +6x = 16

x^2 + 6x - 16 = 0 .... where a= 1, b = 6, c = -16

x = (-b +/- √(b^2-4ac)) / 2a

x = (-(6) +/- √((6)^2 -4(1)(-16))) / 2(1)

x = (-6 +/- √(36 -(-64))) / 2

x = (-6 +/- √100) / 2

x = (-6 +/- 10) / 2

x = 2(-3 +/- 5) / 2

x = -3 +/- 5

x1 = -3 - 5

x1 = -8

x2 = -3 + 5

x2 = 2

THEREFORE x = -8, 2






Solving by factoring by grouping

x^2 + 6x - 16 = 0

x^2 + 8x - 2x - 16 = 0

x(x+8) - 2(x+8) = 0

(x+8)(x-2) = 0

(x+8)=0 and (x-2)=0

x+8=0 and x-2=0

x= -8 and x = 2

THEREFORE x = -8, 2

Put the equation into standard form y = ax^2 +bx +c


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

Root #1
= {-6 + sqrt[36 - 4(1)(-16)]} / 2(1) = {-6 + sqrt[100]}/2 = {-6 + 10}/2 = 4/2 = 2

Root #2
= {-6 - 10}/2 = -16/2 = -8


Astrobuf

Set it equal to zero, them use the quadratic formula. So...

Ax^2+bx+c=0

(-b +or- (b^2 -4ac)^1/2 (square root))/2a

(-6 +or- (36-32)^1/2)/2= -2 or -4

So the two solutions are -2 and -4. :)

move the 16 to the other side
x^2 +6x - 16 =0
simplify
(x+8)(x-2) = 0
x=-8 x=2

x^2 + 6x = 16

x^2 + 6x - 16 = 0

(x + 8)(x - 2)

x^2+6x-16=0
x^2+8x-2x-16=0
x(x+8)-2(x+8)=0
(x+8)(x-2)=0
x=-8 or x=2