What are the solutions to the equation x^2-16x+32=0 ?

The solutions to the equation x^2-16x+32=0 are x=4(2+sqrt(2)),x=4(2-sqrt(2))

Find the zeros of x^2-16x+32=0

The zeros of x^2-16x+32=0 are x=4(2+sqrt(2)),x=4(2-sqrt(2))

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x^{2} - 16 x + 32 = 0

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

Decimal Notation

x = 13.65685 \dots, x = 2.34314 \dots

Solution steps

x^{2} - 16 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 32 = 82

x_{1, 2} = \frac{- ( - 16 ) \pm 8 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8 2}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 8 2}{2 \cdot 1}

x = \frac{- ( - 16 ) + 8 2}{2 \cdot 1} : 4 (2 + 2)

x = \frac{- ( - 16 ) - 8 2}{2 \cdot 1} : 4 (2 - 2)

The solutions to the quadratic equation are:

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

45 x^{2} + 5 = 0

8 x^{2} - x - 1 = 0

7 x^{2} + 7 x - 42 = 0

2 (x - 3)^{2} + 4 = 102

4 (x + 1) - 2 (x - 2) = x^{2} - x - 2

What are the solutions to the equation x^2-16x+32=0 ?
The solutions to the equation x^2-16x+32=0 are x=4(2+sqrt(2)),x=4(2-sqrt(2))
Find the zeros of x^2-16x+32=0
The zeros of x^2-16x+32=0 are x=4(2+sqrt(2)),x=4(2-sqrt(2))