What are the solutions to the equation 3(x+2)(x-2)=(x-4)2+8x ?

The solutions to the equation 3(x+2)(x-2)=(x-4)2+8x are x=(5+sqrt(37))/3 ,x=(5-sqrt(37))/3

Find the zeros of 3(x+2)(x-2)=(x-4)2+8x

The zeros of 3(x+2)(x-2)=(x-4)2+8x are x=(5+sqrt(37))/3 ,x=(5-sqrt(37))/3

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3 (x + 2) (x - 2) = (x - 4) 2 + 8 x

x = \frac{5 + 37}{3}, x = \frac{5 - 37}{3}

Decimal Notation

x = 3.69425 \dots, x = - 0.36092 \dots

Solution steps

3 (x + 2) (x - 2) = (x - 4) \cdot 2 + 8 x

Expand

3 (x + 2) (x - 2) : 3 x^{2} - 12

Expand

(x - 4) \cdot 2 + 8 x : 10 x - 8

3 x^{2} - 12 = 10 x - 8

Move

8

to the left side

3 x^{2} - 4 = 10 x

Move

10 x

to the left side

3 x^{2} - 4 - 10 x = 0

Write in the standard form

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

3 x^{2} - 10 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 (- 4) = 237

x_{1, 2} = \frac{- ( - 10 ) \pm 2 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 37}{2 \cdot 3}, x_{2} = \frac{- ( - 10 ) - 2 37}{2 \cdot 3}

x = \frac{- ( - 10 ) + 2 37}{2 \cdot 3} : \frac{5 + 37}{3}

x = \frac{- ( - 10 ) - 2 37}{2 \cdot 3} : \frac{5 - 37}{3}

The solutions to the quadratic equation are:

x = \frac{5 + 37}{3}, x = \frac{5 - 37}{3}

3 x^{2} + 27 x + 54 = - \frac{3 x ^{2}}{2} - \frac{27 x}{2} - 27

t, 48 = 8 (t) + 16 (t)^{2}

x^{2} + \frac{8}{2} x + \frac{15}{4} = 0

p^{2} + 9 p = 0

3 x^{2} + 2 x - 8 = 0

What are the solutions to the equation 3(x+2)(x-2)=(x-4)2+8x ?
The solutions to the equation 3(x+2)(x-2)=(x-4)2+8x are x=(5+sqrt(37))/3 ,x=(5-sqrt(37))/3
Find the zeros of 3(x+2)(x-2)=(x-4)2+8x
The zeros of 3(x+2)(x-2)=(x-4)2+8x are x=(5+sqrt(37))/3 ,x=(5-sqrt(37))/3