What are the solutions to the equation 4.9t^2+9t+3=0 ?

The solutions to the equation 4.9t^2+9t+3=0 are t=(-45+sqrt(555))/(49),t=-(45+sqrt(555))/(49)

Find the zeros of 4.9t^2+9t+3=0

The zeros of 4.9t^2+9t+3=0 are t=(-45+sqrt(555))/(49),t=-(45+sqrt(555))/(49)

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4.9 t^{2} + 9 t + 3 = 0

t = \frac{- 45 + 555}{49}, t = - \frac{45 + 555}{49}

Decimal Notation

t = - 0.43758 \dots, t = - 1.39915 \dots

Solution steps

4.9 t^{2} + 9 t + 3 = 0

Multiply both sides by

10

49 t^{2} + 90 t + 30 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 90 \pm 9 0 ^{2} - 4 \cdot 49 \cdot 30}{2 \cdot 49}

9 0^{2} - 4 \cdot 49 \cdot 30 = 2555

t_{1, 2} = \frac{- 90 \pm 2 555}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- 90 + 2 555}{2 \cdot 49}, t_{2} = \frac{- 90 - 2 555}{2 \cdot 49}

t = \frac{- 90 + 2 555}{2 \cdot 49} : \frac{- 45 + 555}{49}

t = \frac{- 90 - 2 555}{2 \cdot 49} : - \frac{45 + 555}{49}

The solutions to the quadratic equation are:

t = \frac{- 45 + 555}{49}, t = - \frac{45 + 555}{49}

3 x^{2} + 27 = 12

9 x^{2} + 8 x - 4 = 0

x^{2} + 13 x - 10 = 0

x, f = x^{2} y^{2} z^{2}

(\frac{3}{5})^{2} + x^{2} = 1

What are the solutions to the equation 4.9t^2+9t+3=0 ?
The solutions to the equation 4.9t^2+9t+3=0 are t=(-45+sqrt(555))/(49),t=-(45+sqrt(555))/(49)
Find the zeros of 4.9t^2+9t+3=0
The zeros of 4.9t^2+9t+3=0 are t=(-45+sqrt(555))/(49),t=-(45+sqrt(555))/(49)