What are the solutions to the equation 2t^2+3t-1=0 ?

The solutions to the equation 2t^2+3t-1=0 are t=(-3+sqrt(17))/4 ,t=(-3-sqrt(17))/4

Find the zeros of 2t^2+3t-1=0

The zeros of 2t^2+3t-1=0 are t=(-3+sqrt(17))/4 ,t=(-3-sqrt(17))/4

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2 t^{2} + 3 t - 1 = 0

t = \frac{- 3 + 17}{4}, t = \frac{- 3 - 17}{4}

Decimal Notation

t = 0.28077 \dots, t = - 1.78077 \dots

Solution steps

2 t^{2} + 3 t - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

t = \frac{- 3 + 17}{2 \cdot 2} : \frac{- 3 + 17}{4}

t = \frac{- 3 - 17}{2 \cdot 2} : \frac{- 3 - 17}{4}

The solutions to the quadratic equation are:

t = \frac{- 3 + 17}{4}, t = \frac{- 3 - 17}{4}

(x^{2} + 3 x + 2) 2 - 8 (x^{2} + 3 x) = 4

2 x^{2} + 3 = 49

2 x^{2} + 3 = 2 x

7 x^{2} - 10 = x^{2} - 4

x (3 x) = 550

What are the solutions to the equation 2t^2+3t-1=0 ?
The solutions to the equation 2t^2+3t-1=0 are t=(-3+sqrt(17))/4 ,t=(-3-sqrt(17))/4
Find the zeros of 2t^2+3t-1=0
The zeros of 2t^2+3t-1=0 are t=(-3+sqrt(17))/4 ,t=(-3-sqrt(17))/4