What are the solutions to the equation (x-4)^2+4(x-4)-12=0 ?

The solutions to the equation (x-4)^2+4(x-4)-12=0 are x=6,x=-2

Find the zeros of (x-4)^2+4(x-4)-12=0

The zeros of (x-4)^2+4(x-4)-12=0 are x=6,x=-2

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(x - 4)^{2} + 4 (x - 4) - 12 = 0

x = 6, x = - 2

Solution steps

(x - 4)^{2} + 4 (x - 4) - 12 = 0

Expand

(x - 4)^{2} + 4 (x - 4) - 12 : x^{2} - 4 x - 12

x^{2} - 4 x - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

x = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

x = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 6, x = - 2

x^{2} = x - 6

X + 42 < 45

23 + 312

- 16 (50)^{2} + 70 (50) + 5 = 0

10 ∣ 7 x + 3 ∣ = 0

What are the solutions to the equation (x-4)^2+4(x-4)-12=0 ?
The solutions to the equation (x-4)^2+4(x-4)-12=0 are x=6,x=-2
Find the zeros of (x-4)^2+4(x-4)-12=0
The zeros of (x-4)^2+4(x-4)-12=0 are x=6,x=-2