What are the solutions to the equation 16t^2+80t+32=0 ?

The solutions to the equation 16t^2+80t+32=0 are t=(-5+sqrt(17))/2 ,t=-(5+sqrt(17))/2

Find the zeros of 16t^2+80t+32=0

The zeros of 16t^2+80t+32=0 are t=(-5+sqrt(17))/2 ,t=-(5+sqrt(17))/2

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16 t^{2} + 80 t + 32 = 0

t = \frac{- 5 + 17}{2}, t = - \frac{5 + 17}{2}

Decimal Notation

t = - 0.43844 \dots, t = - 4.56155 \dots

Solution steps

16 t^{2} + 80 t + 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 16 \cdot 32}{2 \cdot 16}

8 0^{2} - 4 \cdot 16 \cdot 32 = 1617

t_{1, 2} = \frac{- 80 \pm 16 17}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 80 + 16 17}{2 \cdot 16}, t_{2} = \frac{- 80 - 16 17}{2 \cdot 16}

t = \frac{- 80 + 16 17}{2 \cdot 16} : \frac{- 5 + 17}{2}

t = \frac{- 80 - 16 17}{2 \cdot 16} : - \frac{5 + 17}{2}

The solutions to the quadratic equation are:

t = \frac{- 5 + 17}{2}, t = - \frac{5 + 17}{2}

- 2 x^{2} + 8

(x - 4) (x + 1) (x - 9) \geq 0

lo g_{3} (x + 3) < 3

\frac{4 a}{3 b c} + \frac{15 b}{5 a c}

\frac{4}{x ^{2} - 1} > 1

What are the solutions to the equation 16t^2+80t+32=0 ?
The solutions to the equation 16t^2+80t+32=0 are t=(-5+sqrt(17))/2 ,t=-(5+sqrt(17))/2
Find the zeros of 16t^2+80t+32=0
The zeros of 16t^2+80t+32=0 are t=(-5+sqrt(17))/2 ,t=-(5+sqrt(17))/2