What are the solutions to the equation 0.2x^2+x-140=0 ?

The solutions to the equation 0.2x^2+x-140=0 are x=(5(sqrt(113)-1))/2 ,x=-(5(1+sqrt(113)))/2

Find the zeros of 0.2x^2+x-140=0

The zeros of 0.2x^2+x-140=0 are x=(5(sqrt(113)-1))/2 ,x=-(5(1+sqrt(113)))/2

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0.2 x^{2} + x - 140 = 0

x = \frac{5 ( 113 - 1 )}{2}, x = - \frac{5 ( 1 + 113 )}{2}

Decimal Notation

x = 24.07536 \dots, x = - 29.07536 \dots

Solution steps

0.2 x^{2} + x - 140 = 0

Multiply both sides by

10

2 x^{2} + 10 x - 1400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 1400 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 1400) = 10113

x_{1, 2} = \frac{- 10 \pm 10 113}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 10 113}{2 \cdot 2}, x_{2} = \frac{- 10 - 10 113}{2 \cdot 2}

x = \frac{- 10 + 10 113}{2 \cdot 2} : \frac{5 ( 113 - 1 )}{2}

x = \frac{- 10 - 10 113}{2 \cdot 2} : - \frac{5 ( 1 + 113 )}{2}

The solutions to the quadratic equation are:

x = \frac{5 ( 113 - 1 )}{2}, x = - \frac{5 ( 1 + 113 )}{2}

2 x^{2} - 33 = 39

0.1 x^{2} + x - 100 = 0

- 8 r^{2} = - 512

(z + 8)^{2} = - 4

x, a (1 - \frac{x}{k}) x - b x y = 0

What are the solutions to the equation 0.2x^2+x-140=0 ?
The solutions to the equation 0.2x^2+x-140=0 are x=(5(sqrt(113)-1))/2 ,x=-(5(1+sqrt(113)))/2
Find the zeros of 0.2x^2+x-140=0
The zeros of 0.2x^2+x-140=0 are x=(5(sqrt(113)-1))/2 ,x=-(5(1+sqrt(113)))/2