What are the solutions to the equation 3t^2+11t-14=0 ?

The solutions to the equation 3t^2+11t-14=0 are t=1,t=-14/3

Find the zeros of 3t^2+11t-14=0

The zeros of 3t^2+11t-14=0 are t=1,t=-14/3

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3 t^{2} + 11 t - 14 = 0

t = 1, t = - \frac{14}{3}

Decimal Notation

t = 1, t = - 4.66666 \dots

Solution steps

3 t^{2} + 11 t - 14 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 ( - 14 )}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 (- 14) = 17

t_{1, 2} = \frac{- 11 \pm 17}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- 11 + 17}{2 \cdot 3}, t_{2} = \frac{- 11 - 17}{2 \cdot 3}

t = \frac{- 11 + 17}{2 \cdot 3} : 1

t = \frac{- 11 - 17}{2 \cdot 3} : - \frac{14}{3}

The solutions to the quadratic equation are:

t = 1, t = - \frac{14}{3}

2 (\frac{1}{4})^{- 2}

- 12 + r = 7

1.2 5^{- 0.2}

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

x^{2} - 8 x + 16 = 25

What are the solutions to the equation 3t^2+11t-14=0 ?
The solutions to the equation 3t^2+11t-14=0 are t=1,t=-14/3
Find the zeros of 3t^2+11t-14=0
The zeros of 3t^2+11t-14=0 are t=1,t=-14/3