What are the solutions to the equation t^2+20t-4=0 ?

The solutions to the equation t^2+20t-4=0 are t=2(sqrt(26)-5),t=-2(5+sqrt(26))

Find the zeros of t^2+20t-4=0

The zeros of t^2+20t-4=0 are t=2(sqrt(26)-5),t=-2(5+sqrt(26))

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t^{2} + 20 t - 4 = 0

t = 2 (26 - 5), t = - 2 (5 + 26)

Decimal Notation

t = 0.19803 \dots, t = - 20.19803 \dots

Solution steps

t^{2} + 20 t - 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 4) = 426

t_{1, 2} = \frac{- 20 \pm 4 26}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 20 + 4 26}{2 \cdot 1}, t_{2} = \frac{- 20 - 4 26}{2 \cdot 1}

t = \frac{- 20 + 4 26}{2 \cdot 1} : 2 (26 - 5)

t = \frac{- 20 - 4 26}{2 \cdot 1} : - 2 (5 + 26)

The solutions to the quadratic equation are:

t = 2 (26 - 5), t = - 2 (5 + 26)

0 = - 16 t^{2} + 96 t

7 - 2 x \geq 15

25 \cdot 5

- 3 x^{2} + 4 x = 0

2 x - \frac{1 - 3 x}{10} - \frac{2}{3} = 2 (x - 3) + \frac{1}{5}

What are the solutions to the equation t^2+20t-4=0 ?
The solutions to the equation t^2+20t-4=0 are t=2(sqrt(26)-5),t=-2(5+sqrt(26))
Find the zeros of t^2+20t-4=0
The zeros of t^2+20t-4=0 are t=2(sqrt(26)-5),t=-2(5+sqrt(26))