What are the solutions to the equation 4.9t^2+8t-25=0 ?

The solutions to the equation 4.9t^2+8t-25=0 are t=(5(sqrt(554)-8))/(49),t=-(5(8+sqrt(554)))/(49)

Find the zeros of 4.9t^2+8t-25=0

The zeros of 4.9t^2+8t-25=0 are t=(5(sqrt(554)-8))/(49),t=-(5(8+sqrt(554)))/(49)

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4.9 t^{2} + 8 t - 25 = 0

t = \frac{5 ( 554 - 8 )}{49}, t = - \frac{5 ( 8 + 554 )}{49}

Decimal Notation

t = 1.58542 \dots, t = - 3.21808 \dots

Solution steps

4.9 t^{2} + 8 t - 25 = 0

Multiply both sides by

10

49 t^{2} + 80 t - 250 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 49 ( - 250 )}{2 \cdot 49}

8 0^{2} - 4 \cdot 49 (- 250) = 10554

t_{1, 2} = \frac{- 80 \pm 10 554}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- 80 + 10 554}{2 \cdot 49}, t_{2} = \frac{- 80 - 10 554}{2 \cdot 49}

t = \frac{- 80 + 10 554}{2 \cdot 49} : \frac{5 ( 554 - 8 )}{49}

t = \frac{- 80 - 10 554}{2 \cdot 49} : - \frac{5 ( 8 + 554 )}{49}

The solutions to the quadratic equation are:

t = \frac{5 ( 554 - 8 )}{49}, t = - \frac{5 ( 8 + 554 )}{49}

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λ^{2} - 4 λ + 3 = 0

What are the solutions to the equation 4.9t^2+8t-25=0 ?
The solutions to the equation 4.9t^2+8t-25=0 are t=(5(sqrt(554)-8))/(49),t=-(5(8+sqrt(554)))/(49)
Find the zeros of 4.9t^2+8t-25=0
The zeros of 4.9t^2+8t-25=0 are t=(5(sqrt(554)-8))/(49),t=-(5(8+sqrt(554)))/(49)