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

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

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

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

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

t = \frac{- 1 + 3}{2}, t = - \frac{1 + 3}{2}

Decimal

t = 0.36602 \dots, t = - 1.36602 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 1) = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = \frac{- 1 + 3}{2}, t = - \frac{1 + 3}{2}

x^{2} + 3 x + 1

4 X^{2} + 5 X - 51 = 0

4 x^{2} - 20 x = - 45

y, x^{2} y^{3} = 1

x^{2} + 5 x - 6 = 0

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