What are the solutions to the equation solvefor x,x^2-4x+2^t=0 ?

The solutions to the equation solvefor x,x^2-4x+2^t=0 are x=2+sqrt(4-2^t),x=2-sqrt(4-2^t)

Find the zeros of solvefor x,x^2-4x+2^t=0

The zeros of solvefor x,x^2-4x+2^t=0 are x=2+sqrt(4-2^t),x=2-sqrt(4-2^t)

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solve for

x, x^{2} - 4 x + 2^{t} = 0

x = 2 + 4 - 2^{t}, x = 2 - 4 - 2^{t}

Solution steps

x^{2} - 4 x + 2^{t} = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 2^{t} : 2 4 - 2^{t}

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 4 - 2 ^{t}}{2 \cdot 1} : 2 + 4 - 2^{t}

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

The solutions to the quadratic equation are:

x = 2 + 4 - 2^{t}, x = 2 - 4 - 2^{t}

x, x^{2} + y^{2} + 2 x = - 1

2 d^{2} = 8 d + 90

2 x^{2} + 3 x - 16 = 4

v^{2} = 47

6 x^{2} = - x + 1

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