What are the solutions to the equation 4.9t^2+294t-2450=0 ?

The solutions to the equation 4.9t^2+294t-2450=0 are t=10(sqrt(14)-3),t=-10(3+sqrt(14))

Find the zeros of 4.9t^2+294t-2450=0

The zeros of 4.9t^2+294t-2450=0 are t=10(sqrt(14)-3),t=-10(3+sqrt(14))

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4.9 t^{2} + 294 t - 2450 = 0

t = 10 (14 - 3), t = - 10 (3 + 14)

Decimal Notation

t = 7.41657 \dots, t = - 67.41657 \dots

Solution steps

4.9 t^{2} + 294 t - 2450 = 0

Multiply both sides by

10

49 t^{2} + 2940 t - 24500 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2940 \pm 294 0 ^{2} - 4 \cdot 49 ( - 24500 )}{2 \cdot 49}

294 0^{2} - 4 \cdot 49 (- 24500) = 98014

t_{1, 2} = \frac{- 2940 \pm 980 14}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- 2940 + 980 14}{2 \cdot 49}, t_{2} = \frac{- 2940 - 980 14}{2 \cdot 49}

t = \frac{- 2940 + 980 14}{2 \cdot 49} : 10 (14 - 3)

t = \frac{- 2940 - 980 14}{2 \cdot 49} : - 10 (3 + 14)

The solutions to the quadratic equation are:

t = 10 (14 - 3), t = - 10 (3 + 14)

x, x^{2} - 25 m^{2} = 0

x^{2} - (c + 6) \cdot x + 2 (2 c - 1) = 0

(k + 1) \cdot x^{2} - 2 (k - 1) \cdot x + 1 = 0

y, u = x^{3} - 3 x \cdot y^{2} + 3 x^{2} - 2 y^{2} + 2 x + 1

x, z = 8 - x^{2} - y^{2}

What are the solutions to the equation 4.9t^2+294t-2450=0 ?
The solutions to the equation 4.9t^2+294t-2450=0 are t=10(sqrt(14)-3),t=-10(3+sqrt(14))
Find the zeros of 4.9t^2+294t-2450=0
The zeros of 4.9t^2+294t-2450=0 are t=10(sqrt(14)-3),t=-10(3+sqrt(14))