What are the solutions to the equation x^2-32x+64=0 ?

The solutions to the equation x^2-32x+64=0 are x=8(2+sqrt(3)),x=8(2-sqrt(3))

Find the zeros of x^2-32x+64=0

The zeros of x^2-32x+64=0 are x=8(2+sqrt(3)),x=8(2-sqrt(3))

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x^{2} - 32 x + 64 = 0

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

Decimal Notation

x = 29.85640 \dots, x = 2.14359 \dots

Solution steps

x^{2} - 32 x + 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 1 \cdot 64}{2 \cdot 1}

(- 32)^{2} - 4 \cdot 1 \cdot 64 = 163

x_{1, 2} = \frac{- ( - 32 ) \pm 16 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 16 3}{2 \cdot 1}, x_{2} = \frac{- ( - 32 ) - 16 3}{2 \cdot 1}

x = \frac{- ( - 32 ) + 16 3}{2 \cdot 1} : 8 (2 + 3)

x = \frac{- ( - 32 ) - 16 3}{2 \cdot 1} : 8 (2 - 3)

The solutions to the quadratic equation are:

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

- 3 x^{2} - 7 + 2 x = - 11

4 x = 6 x^{2}

3 x^{2} - 5 x + c = 0

g^{2} = 6 0^{2} + 1 5^{2}

9 z = 10 z^{2}

What are the solutions to the equation x^2-32x+64=0 ?
The solutions to the equation x^2-32x+64=0 are x=8(2+sqrt(3)),x=8(2-sqrt(3))
Find the zeros of x^2-32x+64=0
The zeros of x^2-32x+64=0 are x=8(2+sqrt(3)),x=8(2-sqrt(3))