What are the solutions to the equation 5t^2+3t-7=0 ?

The solutions to the equation 5t^2+3t-7=0 are t=(-3+sqrt(149))/(10),t=(-3-sqrt(149))/(10)

Find the zeros of 5t^2+3t-7=0

The zeros of 5t^2+3t-7=0 are t=(-3+sqrt(149))/(10),t=(-3-sqrt(149))/(10)

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5 t^{2} + 3 t - 7 = 0

t = \frac{- 3 + 149}{10}, t = \frac{- 3 - 149}{10}

Solution steps

5 t^{2} + 3 t - 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 7) = 149

t_{1, 2} = \frac{- 3 \pm 149}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 3 + 149}{2 \cdot 5}, t_{2} = \frac{- 3 - 149}{2 \cdot 5}

t = \frac{- 3 + 149}{2 \cdot 5} : \frac{- 3 + 149}{10}

t = \frac{- 3 - 149}{2 \cdot 5} : \frac{- 3 - 149}{10}

The solutions to the quadratic equation are:

t = \frac{- 3 + 149}{10}, t = \frac{- 3 - 149}{10}

\frac{16}{3} + \frac{16}{15}

∣ 2 x - 3 ∣ = ∣ 5 x + 4 ∣

\frac{2 x - 1}{x - 3} > 1

x^{2} + 3 x = 5

∣ 5 x + 4 ∣ \leq 7

What are the solutions to the equation 5t^2+3t-7=0 ?
The solutions to the equation 5t^2+3t-7=0 are t=(-3+sqrt(149))/(10),t=(-3-sqrt(149))/(10)
Find the zeros of 5t^2+3t-7=0
The zeros of 5t^2+3t-7=0 are t=(-3+sqrt(149))/(10),t=(-3-sqrt(149))/(10)