What are the solutions to the equation 3.9=-0.5x^2+0.9x+3.8 ?

The solutions to the equation 3.9=-0.5x^2+0.9x+3.8 are x=(9-sqrt(61))/(10),x=(9+sqrt(61))/(10)

Find the zeros of 3.9=-0.5x^2+0.9x+3.8

The zeros of 3.9=-0.5x^2+0.9x+3.8 are x=(9-sqrt(61))/(10),x=(9+sqrt(61))/(10)

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3.9 = - 0.5 x^{2} + 0.9 x + 3.8

x = \frac{9 - 61}{10}, x = \frac{9 + 61}{10}

Solution steps

3.9 = - 0.5 x^{2} + 0.9 x + 3.8

Multiply both sides by

10

39 = - 5 x^{2} + 9 x + 38

Switch sides

- 5 x^{2} + 9 x + 38 = 39

Move

39

to the left side

- 5 x^{2} + 9 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 5 ) ( - 1 )}{2 ( - 5 )}

9^{2} - 4 (- 5) (- 1) = 61

x_{1, 2} = \frac{- 9 \pm 61}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 9 + 61}{2 ( - 5 )}, x_{2} = \frac{- 9 - 61}{2 ( - 5 )}

x = \frac{- 9 + 61}{2 ( - 5 )} : \frac{9 - 61}{10}

x = \frac{- 9 - 61}{2 ( - 5 )} : \frac{9 + 61}{10}

The solutions to the quadratic equation are:

x = \frac{9 - 61}{10}, x = \frac{9 + 61}{10}

q^{2} + 40 q + 400 = 36

\frac{5 ( x - 8 )}{6} = - 7

- \frac{4}{3} m \leq \frac{25}{12}

2 t + 8 = - 10

- x^{2} - x + 12

What are the solutions to the equation 3.9=-0.5x^2+0.9x+3.8 ?
The solutions to the equation 3.9=-0.5x^2+0.9x+3.8 are x=(9-sqrt(61))/(10),x=(9+sqrt(61))/(10)
Find the zeros of 3.9=-0.5x^2+0.9x+3.8
The zeros of 3.9=-0.5x^2+0.9x+3.8 are x=(9-sqrt(61))/(10),x=(9+sqrt(61))/(10)