What are the solutions to the equation roots 27t^2+6t+1t ?

The solutions to the equation roots 27t^2+6t+1t are t=0,t=-7/27

Find the zeros of roots 27t^2+6t+1t

The zeros of roots 27t^2+6t+1t are t=0,t=-7/27

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roots

27 t^{2} + 6 t + 1 t

t = 0, t = - \frac{7}{27}

Solution steps

27 t^{2} + 6 t + 1 \cdot t

The roots are the intercepts with the x-axis

(y = 0)

27 t^{2} + 6 t + 1 \cdot t = 0

Refine

27 t^{2} + 7 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 27 \cdot 0}{2 \cdot 27}

7^{2} - 4 \cdot 27 \cdot 0 = 7

t_{1, 2} = \frac{- 7 \pm 7}{2 \cdot 27}

Separate the solutions

t_{1} = \frac{- 7 + 7}{2 \cdot 27}, t_{2} = \frac{- 7 - 7}{2 \cdot 27}

t = \frac{- 7 + 7}{2 \cdot 27} : 0

t = \frac{- 7 - 7}{2 \cdot 27} : - \frac{7}{27}

The solutions to the quadratic equation are:

t = 0, t = - \frac{7}{27}

3 p^{2} = - 8 p - 5

y, x^{2} + 6 x y + 12 y^{2} = 28

2 y^{2} - y + 1 = 0

2 x^{2} + 5 x - 3

3 x^{2} - 2 x + 6

What are the solutions to the equation roots 27t^2+6t+1t ?
The solutions to the equation roots 27t^2+6t+1t are t=0,t=-7/27
Find the zeros of roots 27t^2+6t+1t
The zeros of roots 27t^2+6t+1t are t=0,t=-7/27