What are the solutions to the equation x^2-40x-800=0 ?

The solutions to the equation x^2-40x-800=0 are x=20(1+sqrt(3)),x=20(1-sqrt(3))

Find the zeros of x^2-40x-800=0

The zeros of x^2-40x-800=0 are x=20(1+sqrt(3)),x=20(1-sqrt(3))

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x^{2} - 40 x - 800 = 0

x = 20 (1 + 3), x = 20 (1 - 3)

Decimal Notation

x = 54.64101 \dots, x = - 14.64101 \dots

Solution steps

x^{2} - 40 x - 800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 1 \cdot ( - 800 )}{2 \cdot 1}

(- 40)^{2} - 4 \cdot 1 \cdot (- 800) = 403

x_{1, 2} = \frac{- ( - 40 ) \pm 40 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 40 3}{2 \cdot 1}, x_{2} = \frac{- ( - 40 ) - 40 3}{2 \cdot 1}

x = \frac{- ( - 40 ) + 40 3}{2 \cdot 1} : 20 (1 + 3)

x = \frac{- ( - 40 ) - 40 3}{2 \cdot 1} : 20 (1 - 3)

The solutions to the quadratic equation are:

x = 20 (1 + 3), x = 20 (1 - 3)

6 x^{2} + 120 x + 2070 = 0

x, x^{2} + 2 x y + y^{2} + 2 x - 4 y + 5 = 0

6 x^{2} = 12 + x

x^{2} = - 16 x

x^{2} = - 14 x

What are the solutions to the equation x^2-40x-800=0 ?
The solutions to the equation x^2-40x-800=0 are x=20(1+sqrt(3)),x=20(1-sqrt(3))
Find the zeros of x^2-40x-800=0
The zeros of x^2-40x-800=0 are x=20(1+sqrt(3)),x=20(1-sqrt(3))