What are the solutions to the equation x^2-8x-32=0 ?

The solutions to the equation x^2-8x-32=0 are x=4(1+sqrt(3)),x=4(1-sqrt(3))

Find the zeros of x^2-8x-32=0

The zeros of x^2-8x-32=0 are x=4(1+sqrt(3)),x=4(1-sqrt(3))

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x^{2} - 8 x - 32 = 0

x = 4 (1 + 3), x = 4 (1 - 3)

Decimal Notation

x = 10.92820 \dots, x = - 2.92820 \dots

Solution steps

x^{2} - 8 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 32) = 83

x_{1, 2} = \frac{- ( - 8 ) \pm 8 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 8 3}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 8 3}{2 \cdot 1}

x = \frac{- ( - 8 ) + 8 3}{2 \cdot 1} : 4 (1 + 3)

x = \frac{- ( - 8 ) - 8 3}{2 \cdot 1} : 4 (1 - 3)

The solutions to the quadratic equation are:

x = 4 (1 + 3), x = 4 (1 - 3)

∣ x + 4 ∣ = ∣ 5 x + 2 ∣

- 7112 + 7

x - \frac{2}{7} = 5

\frac{p}{2.5} - 2.8 = - 3.44

∣ 2 x - 5 ∣ \leq 13

What are the solutions to the equation x^2-8x-32=0 ?
The solutions to the equation x^2-8x-32=0 are x=4(1+sqrt(3)),x=4(1-sqrt(3))
Find the zeros of x^2-8x-32=0
The zeros of x^2-8x-32=0 are x=4(1+sqrt(3)),x=4(1-sqrt(3))