What are the solutions to the equation -0.1x^2+0.1x=-0.03 ?

The solutions to the equation -0.1x^2+0.1x=-0.03 are x=-(-5+sqrt(55))/(10),x=(5+sqrt(55))/(10)

Find the zeros of -0.1x^2+0.1x=-0.03

The zeros of -0.1x^2+0.1x=-0.03 are x=-(-5+sqrt(55))/(10),x=(5+sqrt(55))/(10)

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- 0.1 x^{2} + 0.1 x = - 0.03

x = - \frac{- 5 + 55}{10}, x = \frac{5 + 55}{10}

Decimal Notation

x = - 0.24161 \dots, x = 1.24161 \dots

Solution steps

- 0.1 x^{2} + 0.1 x = - 0.03

Multiply both sides by

100

- 10 x^{2} + 10 x = - 3

Move

3

to the left side

- 10 x^{2} + 10 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 10 ) \cdot 3}{2 ( - 10 )}

1 0^{2} - 4 (- 10) \cdot 3 = 255

x_{1, 2} = \frac{- 10 \pm 2 55}{2 ( - 10 )}

Separate the solutions

x_{1} = \frac{- 10 + 2 55}{2 ( - 10 )}, x_{2} = \frac{- 10 - 2 55}{2 ( - 10 )}

x = \frac{- 10 + 2 55}{2 ( - 10 )} : - \frac{- 5 + 55}{10}

x = \frac{- 10 - 2 55}{2 ( - 10 )} : \frac{5 + 55}{10}

The solutions to the quadratic equation are:

x = - \frac{- 5 + 55}{10}, x = \frac{5 + 55}{10}

(x^{3})^{- 2}

x + 3 \geq 0

2 t^{2} + 9 t + 4

lo g_{5} (625)

x^{2} + 4 x > 5

What are the solutions to the equation -0.1x^2+0.1x=-0.03 ?
The solutions to the equation -0.1x^2+0.1x=-0.03 are x=-(-5+sqrt(55))/(10),x=(5+sqrt(55))/(10)
Find the zeros of -0.1x^2+0.1x=-0.03
The zeros of -0.1x^2+0.1x=-0.03 are x=-(-5+sqrt(55))/(10),x=(5+sqrt(55))/(10)