What are the solutions to the equation 0.5x^2+2.1x=0.3 ?

The solutions to the equation 0.5x^2+2.1x=0.3 are x=(-21+sqrt(501))/(10),x=(-21-sqrt(501))/(10)

Find the zeros of 0.5x^2+2.1x=0.3

The zeros of 0.5x^2+2.1x=0.3 are x=(-21+sqrt(501))/(10),x=(-21-sqrt(501))/(10)

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0.5 x^{2} + 2.1 x = 0.3

x = \frac{- 21 + 501}{10}, x = \frac{- 21 - 501}{10}

Decimal Notation

x = 0.13830 \dots, x = - 4.33830 \dots

Solution steps

0.5 x^{2} + 2.1 x = 0.3

Multiply both sides by

10

5 x^{2} + 21 x = 3

Move

3

to the left side

5 x^{2} + 21 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 21 \pm 2 1 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

2 1^{2} - 4 \cdot 5 (- 3) = 501

x_{1, 2} = \frac{- 21 \pm 501}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 21 + 501}{2 \cdot 5}, x_{2} = \frac{- 21 - 501}{2 \cdot 5}

x = \frac{- 21 + 501}{2 \cdot 5} : \frac{- 21 + 501}{10}

x = \frac{- 21 - 501}{2 \cdot 5} : \frac{- 21 - 501}{10}

The solutions to the quadratic equation are:

x = \frac{- 21 + 501}{10}, x = \frac{- 21 - 501}{10}

x^{2} + 2 x + \frac{1}{4} = 0

- 12 x^{2} - 18 x = 0

2 x^{2} + 10 x + 26 = 8

v, f = \frac{m v ^{2}}{r}

z^{2} + 6 z = 10

What are the solutions to the equation 0.5x^2+2.1x=0.3 ?
The solutions to the equation 0.5x^2+2.1x=0.3 are x=(-21+sqrt(501))/(10),x=(-21-sqrt(501))/(10)
Find the zeros of 0.5x^2+2.1x=0.3
The zeros of 0.5x^2+2.1x=0.3 are x=(-21+sqrt(501))/(10),x=(-21-sqrt(501))/(10)