What are the solutions to the equation 10=50+2x-0.0625x^2 ?

The solutions to the equation 10=50+2x-0.0625x^2 are x=-(-2+sqrt(14))/(0.125),x=(2+sqrt(14))/(0.125)

Find the zeros of 10=50+2x-0.0625x^2

The zeros of 10=50+2x-0.0625x^2 are x=-(-2+sqrt(14))/(0.125),x=(2+sqrt(14))/(0.125)

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10 = 50 + 2 x - 0.0625 x^{2}

x = - \frac{- 2 + 14}{0.125}, x = \frac{2 + 14}{0.125}

Decimal Notation

x = - 13.93325 \dots, x = 45.93325 \dots

Solution steps

10 = 50 + 2 x - 0.0625 x^{2}

Switch sides

50 + 2 x - 0.0625 x^{2} = 10

Move

10

to the left side

- 0.0625 x^{2} + 2 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 0.0625 ) \cdot 40}{2 ( - 0.0625 )}

2^{2} - 4 (- 0.0625) \cdot 40 = 14

x_{1, 2} = \frac{- 2 \pm 14}{2 ( - 0.0625 )}

Separate the solutions

x_{1} = \frac{- 2 + 14}{2 ( - 0.0625 )}, x_{2} = \frac{- 2 - 14}{2 ( - 0.0625 )}

x = \frac{- 2 + 14}{2 ( - 0.0625 )} : - \frac{- 2 + 14}{0.125}

x = \frac{- 2 - 14}{2 ( - 0.0625 )} : \frac{2 + 14}{0.125}

The solutions to the quadratic equation are:

x = - \frac{- 2 + 14}{0.125}, x = \frac{2 + 14}{0.125}

5 + 4 x = 3 \cdot (x - 2) + 9

(19 9 m^{4}) (4 10 m^{2})

\frac{a ^{2} b}{a ^{- 3} b ^{- 3}}

- 12 = 4 (x - 5)

5 t - 3 \leq - 38 or 2 t + 1 > 3

What are the solutions to the equation 10=50+2x-0.0625x^2 ?
The solutions to the equation 10=50+2x-0.0625x^2 are x=-(-2+sqrt(14))/(0.125),x=(2+sqrt(14))/(0.125)
Find the zeros of 10=50+2x-0.0625x^2
The zeros of 10=50+2x-0.0625x^2 are x=-(-2+sqrt(14))/(0.125),x=(2+sqrt(14))/(0.125)