What are the solutions to the equation 3(3x-2)=(x+4)(x-4) ?

The solutions to the equation 3(3x-2)=(x+4)(x-4) are x=10,x=-1

Find the zeros of 3(3x-2)=(x+4)(x-4)

The zeros of 3(3x-2)=(x+4)(x-4) are x=10,x=-1

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3 (3 x - 2) = (x + 4) (x - 4)

x = 10, x = - 1

Solution steps

3 (3 x - 2) = (x + 4) (x - 4)

Expand

3 (3 x - 2) : 9 x - 6

Expand

(x + 4) (x - 4) : x^{2} - 16

9 x - 6 = x^{2} - 16

Switch sides

x^{2} - 16 = 9 x - 6

Move

6

to the left side

x^{2} - 10 = 9 x

Move

9 x

to the left side

x^{2} - 10 - 9 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 9 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 10) = 11

x_{1, 2} = \frac{- ( - 9 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 11}{2 \cdot 1}

x = \frac{- ( - 9 ) + 11}{2 \cdot 1} : 10

x = \frac{- ( - 9 ) - 11}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 10, x = - 1

x^{2} - 5 x - 19 = 0

3 x^{2} - 72 = 6 x

X^{2} - 100 = 0

x^{2} + 12 x + 36 = 5

2 x^{2} + 12 x + \frac{35}{2} = 0

What are the solutions to the equation 3(3x-2)=(x+4)(x-4) ?
The solutions to the equation 3(3x-2)=(x+4)(x-4) are x=10,x=-1
Find the zeros of 3(3x-2)=(x+4)(x-4)
The zeros of 3(3x-2)=(x+4)(x-4) are x=10,x=-1