solve 2x^2+12x+18=0?

2x^2 + 12x +18 = 0

(2x + 6) (1x + 3) = 0

(2x +6) = 0 and (1x +3) = 0

2x = -6 and x = -3

x = -3

2x^2 + 12x + 18 = 0
(2x^2 + 12x + 18)/2 = 0/2
x^2 + 6x + 9 = 0
x^2 + 3x + 3x + 9 = 0
(x^2 + 3x) + (3x + 9) = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0

x + 3 = 0
x = -3

∴ x = -3

x² + 6x + 9 = 0
(x + 3)(x + 3) = 0
x = - 3

=(2X+6) (X+3)
= X= -3

Either use quadratic formula or factor.

I'll factor...

Divide everything by 2 first to make it easier...

2 ( x^2 + 6x + 9) = 0

2 ( x + 3) (x + 3 ) = 0

Use zero product property

Solution is x = - 3.

The 2 on the outside doesn't matter because 2 does not equal zero.

The solution of x = -3 is called a double root because the solution of -3 occurred twice.

Good luck to you !

factor out 2 --> 2( x^2 +6x +9 ) = 0
factor the polynomial (x+3) (x+3) = X^2 +6x +9)

therefore 2 (x+3) (X+3) = 0 x = - 3

(2x+6)(x+3)=0
x=-3

make it a good day

2x^2+12x+18=0
4x+12x+18=0
16x+18=0
16x=-18
x=-1.125