What are the solutions to the equation 2t^2+4t-6=0 ?

The solutions to the equation 2t^2+4t-6=0 are t=1,t=-3

Find the zeros of 2t^2+4t-6=0

The zeros of 2t^2+4t-6=0 are t=1,t=-3

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2 t^{2} + 4 t - 6 = 0

t = 1, t = - 3

Solution steps

2 t^{2} + 4 t - 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 6) = 8

t_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 4 + 8}{2 \cdot 2}, t_{2} = \frac{- 4 - 8}{2 \cdot 2}

t = \frac{- 4 + 8}{2 \cdot 2} : 1

t = \frac{- 4 - 8}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

t = 1, t = - 3

x^{2} - \frac{5}{3} x - \frac{2}{3} = 0

a, a^{2} + b^{2} + c^{2} = 121

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

x^{2} + 3 = 51

x^{2} + 6 x + 34

What are the solutions to the equation 2t^2+4t-6=0 ?
The solutions to the equation 2t^2+4t-6=0 are t=1,t=-3
Find the zeros of 2t^2+4t-6=0
The zeros of 2t^2+4t-6=0 are t=1,t=-3