What are the solutions to the equation 0=12+29.4x-4.9x^2 ?

The solutions to the equation 0=12+29.4x-4.9x^2 are x=-(-21+sqrt(561))/7 ,x=(21+sqrt(561))/7

Find the zeros of 0=12+29.4x-4.9x^2

The zeros of 0=12+29.4x-4.9x^2 are x=-(-21+sqrt(561))/7 ,x=(21+sqrt(561))/7

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0 = 12 + 29.4 x - 4.9 x^{2}

x = - \frac{- 21 + 561}{7}, x = \frac{21 + 561}{7}

Decimal Notation

x = - 0.38363 \dots, x = 6.38363 \dots

Solution steps

0 = 12 + 29.4 x - 4.9 x^{2}

Multiply both sides by

10

0 = 120 + 294 x - 49 x^{2}

Switch sides

120 + 294 x - 49 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 49 x^{2} + 294 x + 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 294 \pm 29 4 ^{2} - 4 ( - 49 ) \cdot 120}{2 ( - 49 )}

29 4^{2} - 4 (- 49) \cdot 120 = 14561

x_{1, 2} = \frac{- 294 \pm 14 561}{2 ( - 49 )}

Separate the solutions

x_{1} = \frac{- 294 + 14 561}{2 ( - 49 )}, x_{2} = \frac{- 294 - 14 561}{2 ( - 49 )}

x = \frac{- 294 + 14 561}{2 ( - 49 )} : - \frac{- 21 + 561}{7}

x = \frac{- 294 - 14 561}{2 ( - 49 )} : \frac{21 + 561}{7}

The solutions to the quadratic equation are:

x = - \frac{- 21 + 561}{7}, x = \frac{21 + 561}{7}

x - 2 \geq (x - 2) (2 x^{2} - x - 15)

(\frac{15}{14})^{2} + (\frac{15}{14})^{2} = c^{2}

4 x^{2} = - 2 x + 12

∣ x + 6 ∣ \geq 2

- 1 - \frac{1}{3}

What are the solutions to the equation 0=12+29.4x-4.9x^2 ?
The solutions to the equation 0=12+29.4x-4.9x^2 are x=-(-21+sqrt(561))/7 ,x=(21+sqrt(561))/7
Find the zeros of 0=12+29.4x-4.9x^2
The zeros of 0=12+29.4x-4.9x^2 are x=-(-21+sqrt(561))/7 ,x=(21+sqrt(561))/7