What are the solutions to the equation x^2-4x-160=0 ?

The solutions to the equation x^2-4x-160=0 are x=2(1+sqrt(41)),x=2(1-sqrt(41))

Find the zeros of x^2-4x-160=0

The zeros of x^2-4x-160=0 are x=2(1+sqrt(41)),x=2(1-sqrt(41))

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x^{2} - 4 x - 160 = 0

x = 2 (1 + 41), x = 2 (1 - 41)

Decimal Notation

x = 14.80624 \dots, x = - 10.80624 \dots

Solution steps

x^{2} - 4 x - 160 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 160 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 160) = 441

x_{1, 2} = \frac{- ( - 4 ) \pm 4 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 41}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 41}{2 \cdot 1}

x = \frac{- ( - 4 ) + 4 41}{2 \cdot 1} : 2 (1 + 41)

x = \frac{- ( - 4 ) - 4 41}{2 \cdot 1} : 2 (1 - 41)

The solutions to the quadratic equation are:

x = 2 (1 + 41), x = 2 (1 - 41)

x, 7 x^{2} + 2 y^{2} + 2 x - 7 y + 16 = 0

266 = 0.68 \cdot 100 \cdot X \cdot 1.79 \cdot (7.2 - 0.4 \cdot X)

0.7 x^{2} - 4.2 x - 0.7 = 0

- 4 x^{2} + 225 = 45

0.1 x^{2} + 0.4 x - 0.3 = 0

What are the solutions to the equation x^2-4x-160=0 ?
The solutions to the equation x^2-4x-160=0 are x=2(1+sqrt(41)),x=2(1-sqrt(41))
Find the zeros of x^2-4x-160=0
The zeros of x^2-4x-160=0 are x=2(1+sqrt(41)),x=2(1-sqrt(41))