What are the solutions to the equation t^2+16t-4=0 ?

The solutions to the equation t^2+16t-4=0 are t=2(sqrt(17)-4),t=-2(4+sqrt(17))

Find the zeros of t^2+16t-4=0

The zeros of t^2+16t-4=0 are t=2(sqrt(17)-4),t=-2(4+sqrt(17))

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t^{2} + 16 t - 4 = 0

t = 2 (17 - 4), t = - 2 (4 + 17)

Decimal Notation

t = 0.24621 \dots, t = - 16.24621 \dots

Solution steps

t^{2} + 16 t - 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot (- 4) = 417

t_{1, 2} = \frac{- 16 \pm 4 17}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 16 + 4 17}{2 \cdot 1}, t_{2} = \frac{- 16 - 4 17}{2 \cdot 1}

t = \frac{- 16 + 4 17}{2 \cdot 1} : 2 (17 - 4)

t = \frac{- 16 - 4 17}{2 \cdot 1} : - 2 (4 + 17)

The solutions to the quadratic equation are:

t = 2 (17 - 4), t = - 2 (4 + 17)

\frac{n ( n + 1 )}{2} = 575

x^{2} - 8 x + c

3 x^{2} - 9 x + 4 = 0

\frac{3 ( x ^{2} + 6 )}{2} = 1

4 x^{2} + 16 x + 1 = 0

What are the solutions to the equation t^2+16t-4=0 ?
The solutions to the equation t^2+16t-4=0 are t=2(sqrt(17)-4),t=-2(4+sqrt(17))
Find the zeros of t^2+16t-4=0
The zeros of t^2+16t-4=0 are t=2(sqrt(17)-4),t=-2(4+sqrt(17))