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

The solutions to the equation x^2-x-40=0 are x=(1+sqrt(161))/2 ,x=(1-sqrt(161))/2

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

The zeros of x^2-x-40=0 are x=(1+sqrt(161))/2 ,x=(1-sqrt(161))/2

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x^{2} - x - 40 = 0

x = \frac{1 + 161}{2}, x = \frac{1 - 161}{2}

Decimal

x = 6.84428 \dots, x = - 5.84428 \dots

Solution steps

x^{2} - x - 40 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 40) = 161

x_{1, 2} = \frac{- ( - 1 ) \pm 161}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 161}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 161}{2 \cdot 1}

x = \frac{- ( - 1 ) + 161}{2 \cdot 1} : \frac{1 + 161}{2}

x = \frac{- ( - 1 ) - 161}{2 \cdot 1} : \frac{1 - 161}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 161}{2}, x = \frac{1 - 161}{2}

n^{2} - 3 n - 418 = 0

0 = 8 x - 2 x^{2}

(x - 5)^{2} - 25 = 0

2 x^{2} = 28 - 10 x

x, 2 x^{2} - 7 x + 5 = 0

What are the solutions to the equation x^2-x-40=0 ?
The solutions to the equation x^2-x-40=0 are x=(1+sqrt(161))/2 ,x=(1-sqrt(161))/2
Find the zeros of x^2-x-40=0
The zeros of x^2-x-40=0 are x=(1+sqrt(161))/2 ,x=(1-sqrt(161))/2