What are the solutions to the equation 2t^2+40t+34=0 ?

The solutions to the equation 2t^2+40t+34=0 are t=-10+sqrt(83),t=-10-sqrt(83)

Find the zeros of 2t^2+40t+34=0

The zeros of 2t^2+40t+34=0 are t=-10+sqrt(83),t=-10-sqrt(83)

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2 t^{2} + 40 t + 34 = 0

t = - 10 + 83, t = - 10 - 83

Decimal Notation

t = - 0.88956 \dots, t = - 19.11043 \dots

Solution steps

2 t^{2} + 40 t + 34 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 2 \cdot 34}{2 \cdot 2}

4 0^{2} - 4 \cdot 2 \cdot 34 = 483

t_{1, 2} = \frac{- 40 \pm 4 83}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 40 + 4 83}{2 \cdot 2}, t_{2} = \frac{- 40 - 4 83}{2 \cdot 2}

t = \frac{- 40 + 4 83}{2 \cdot 2} : - 10 + 83

t = \frac{- 40 - 4 83}{2 \cdot 2} : - 10 - 83

The solutions to the quadratic equation are:

t = - 10 + 83, t = - 10 - 83

15 x^{2} - 23 x + 4 = 0

2 x^{2} + 20 x

0.15 y (y - 0.2) = 2 - 0.5 (1 - y)

- 3 x^{2} - 8 x + 7 = 0

2 r^{2} - 11 r + 5 = 0

What are the solutions to the equation 2t^2+40t+34=0 ?
The solutions to the equation 2t^2+40t+34=0 are t=-10+sqrt(83),t=-10-sqrt(83)
Find the zeros of 2t^2+40t+34=0
The zeros of 2t^2+40t+34=0 are t=-10+sqrt(83),t=-10-sqrt(83)