What are the solutions to the equation -5t^2+9t+3=0 ?

The solutions to the equation -5t^2+9t+3=0 are t=-(-9+sqrt(141))/(10),t=(9+sqrt(141))/(10)

Find the zeros of -5t^2+9t+3=0

The zeros of -5t^2+9t+3=0 are t=-(-9+sqrt(141))/(10),t=(9+sqrt(141))/(10)

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- 5 t^{2} + 9 t + 3 = 0

t = - \frac{- 9 + 141}{10}, t = \frac{9 + 141}{10}

Decimal Notation

t = - 0.28743 \dots, t = 2.08743 \dots

Solution steps

- 5 t^{2} + 9 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

9^{2} - 4 (- 5) \cdot 3 = 141

t_{1, 2} = \frac{- 9 \pm 141}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 9 + 141}{2 ( - 5 )}, t_{2} = \frac{- 9 - 141}{2 ( - 5 )}

t = \frac{- 9 + 141}{2 ( - 5 )} : - \frac{- 9 + 141}{10}

t = \frac{- 9 - 141}{2 ( - 5 )} : \frac{9 + 141}{10}

The solutions to the quadratic equation are:

t = - \frac{- 9 + 141}{10}, t = \frac{9 + 141}{10}

2 x^{2} + 20 x - 8 = 0

321.0069 x^{2} - 5130.1778 x + 14682.6031 = 0

x (5 x - 2) = 24

x^{2} - 18 x + 4 = 0

- x^{2} - 12 x + 9 = 0

What are the solutions to the equation -5t^2+9t+3=0 ?
The solutions to the equation -5t^2+9t+3=0 are t=-(-9+sqrt(141))/(10),t=(9+sqrt(141))/(10)
Find the zeros of -5t^2+9t+3=0
The zeros of -5t^2+9t+3=0 are t=-(-9+sqrt(141))/(10),t=(9+sqrt(141))/(10)