What are the solutions to the equation 5t^2+1.3t-0.7=0 ?

The solutions to the equation 5t^2+1.3t-0.7=0 are t=(-13+sqrt(1569))/(100),t=(-13-sqrt(1569))/(100)

Find the zeros of 5t^2+1.3t-0.7=0

The zeros of 5t^2+1.3t-0.7=0 are t=(-13+sqrt(1569))/(100),t=(-13-sqrt(1569))/(100)

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5 t^{2} + 1.3 t - 0.7 = 0

t = \frac{- 13 + 1569}{100}, t = \frac{- 13 - 1569}{100}

Decimal Notation

t = 0.26610 \dots, t = - 0.52610 \dots

Solution steps

5 t^{2} + 1.3 t - 0.7 = 0

Multiply both sides by

10

50 t^{2} + 13 t - 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 50 ( - 7 )}{2 \cdot 50}

1 3^{2} - 4 \cdot 50 (- 7) = 1569

t_{1, 2} = \frac{- 13 \pm 1569}{2 \cdot 50}

Separate the solutions

t_{1} = \frac{- 13 + 1569}{2 \cdot 50}, t_{2} = \frac{- 13 - 1569}{2 \cdot 50}

t = \frac{- 13 + 1569}{2 \cdot 50} : \frac{- 13 + 1569}{100}

t = \frac{- 13 - 1569}{2 \cdot 50} : \frac{- 13 - 1569}{100}

The solutions to the quadratic equation are:

t = \frac{- 13 + 1569}{100}, t = \frac{- 13 - 1569}{100}

y, (10 - x) y^{2} = x^{3}

x^{2} + 5 x - 2 = 5 x + 2

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

s^{2} b + (s + 10) a = 1

1.9 \cdot 1 0^{- 6} x^{2} + 0.013 x + 1 = 0

What are the solutions to the equation 5t^2+1.3t-0.7=0 ?
The solutions to the equation 5t^2+1.3t-0.7=0 are t=(-13+sqrt(1569))/(100),t=(-13-sqrt(1569))/(100)
Find the zeros of 5t^2+1.3t-0.7=0
The zeros of 5t^2+1.3t-0.7=0 are t=(-13+sqrt(1569))/(100),t=(-13-sqrt(1569))/(100)