What are the solutions to the equation 4t^2+11t+2=0 ?

The solutions to the equation 4t^2+11t+2=0 are t=(-11+sqrt(89))/8 ,t=(-11-sqrt(89))/8

Find the zeros of 4t^2+11t+2=0

The zeros of 4t^2+11t+2=0 are t=(-11+sqrt(89))/8 ,t=(-11-sqrt(89))/8

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4 t^{2} + 11 t + 2 = 0

t = \frac{- 11 + 89}{8}, t = \frac{- 11 - 89}{8}

Decimal Notation

t = - 0.19575 \dots, t = - 2.55424 \dots

Solution steps

4 t^{2} + 11 t + 2 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

1 1^{2} - 4 \cdot 4 \cdot 2 = 89

t_{1, 2} = \frac{- 11 \pm 89}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 11 + 89}{2 \cdot 4}, t_{2} = \frac{- 11 - 89}{2 \cdot 4}

t = \frac{- 11 + 89}{2 \cdot 4} : \frac{- 11 + 89}{8}

t = \frac{- 11 - 89}{2 \cdot 4} : \frac{- 11 - 89}{8}

The solutions to the quadratic equation are:

t = \frac{- 11 + 89}{8}, t = \frac{- 11 - 89}{8}

g (- 4) = 3 (- 4)^{3} + 5

2 x + \frac{x}{y}

\frac{7}{12 x} - \frac{y}{6 x ^{2}}

125 n

(r + 4)^{2} = 13

What are the solutions to the equation 4t^2+11t+2=0 ?
The solutions to the equation 4t^2+11t+2=0 are t=(-11+sqrt(89))/8 ,t=(-11-sqrt(89))/8
Find the zeros of 4t^2+11t+2=0
The zeros of 4t^2+11t+2=0 are t=(-11+sqrt(89))/8 ,t=(-11-sqrt(89))/8