What are the solutions to the equation 3x^2+20x+16=6x^2+32x+28 ?

The solutions to the equation 3x^2+20x+16=6x^2+32x+28 are x=-2

Find the zeros of 3x^2+20x+16=6x^2+32x+28

The zeros of 3x^2+20x+16=6x^2+32x+28 are x=-2

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3 x^{2} + 20 x + 16 = 6 x^{2} + 32 x + 28

x = - 2

Solution steps

3 x^{2} + 20 x + 16 = 6 x^{2} + 32 x + 28

Move

28

to the left side

3 x^{2} + 20 x - 12 = 6 x^{2} + 32 x

Move

32 x

to the left side

3 x^{2} - 12 x - 12 = 6 x^{2}

Move

6 x^{2}

to the left side

- 3 x^{2} - 12 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 3 ) ( - 12 )}{2 ( - 3 )}

(- 12)^{2} - 4 (- 3) (- 12) = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 ( - 3 )}

x = \frac{- ( - 12 )}{2 ( - 3 )}

\frac{- ( - 12 )}{2 ( - 3 )} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

(x + 5)^{2} = (5 - x)^{2} + (5 - x)^{2}

(- x + 2) (4 x + 4) = 0

a^{2} - 49 = 0

2 x^{2} + 13 x = - 6

6 x^{2} + 49 x + 77 = 3 x^{2} + 28 + 47

What are the solutions to the equation 3x^2+20x+16=6x^2+32x+28 ?
The solutions to the equation 3x^2+20x+16=6x^2+32x+28 are x=-2
Find the zeros of 3x^2+20x+16=6x^2+32x+28
The zeros of 3x^2+20x+16=6x^2+32x+28 are x=-2