What are the solutions to the equation 0.2x^2+x-0.42=0 ?

The solutions to the equation 0.2x^2+x-0.42=0 are x=(-25+sqrt(835))/(10),x=-(25+sqrt(835))/(10)

Find the zeros of 0.2x^2+x-0.42=0

The zeros of 0.2x^2+x-0.42=0 are x=(-25+sqrt(835))/(10),x=-(25+sqrt(835))/(10)

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0.2 x^{2} + x - 0.42 = 0

x = \frac{- 25 + 835}{10}, x = - \frac{25 + 835}{10}

Decimal Notation

x = 0.38963 \dots, x = - 5.38963 \dots

Solution steps

0.2 x^{2} + x - 0.42 = 0

Multiply both sides by

100

20 x^{2} + 100 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 20 ( - 42 )}{2 \cdot 20}

10 0^{2} - 4 \cdot 20 (- 42) = 4835

x_{1, 2} = \frac{- 100 \pm 4 835}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- 100 + 4 835}{2 \cdot 20}, x_{2} = \frac{- 100 - 4 835}{2 \cdot 20}

x = \frac{- 100 + 4 835}{2 \cdot 20} : \frac{- 25 + 835}{10}

x = \frac{- 100 - 4 835}{2 \cdot 20} : - \frac{25 + 835}{10}

The solutions to the quadratic equation are:

x = \frac{- 25 + 835}{10}, x = - \frac{25 + 835}{10}

x^{2} - 7 x - 15 = 0

2 1^{2} + 7 2^{2} = c^{2}

x^{2} + 3 x - 6 = 4

4 r^{2} - 4 r - 3 = 0

x^{2} - 4 x + 8 = 4

What are the solutions to the equation 0.2x^2+x-0.42=0 ?
The solutions to the equation 0.2x^2+x-0.42=0 are x=(-25+sqrt(835))/(10),x=-(25+sqrt(835))/(10)
Find the zeros of 0.2x^2+x-0.42=0
The zeros of 0.2x^2+x-0.42=0 are x=(-25+sqrt(835))/(10),x=-(25+sqrt(835))/(10)