What are the solutions to the equation 5t^2-14t+6=0 ?

The solutions to the equation 5t^2-14t+6=0 are t=(7+sqrt(19))/5 ,t=(7-sqrt(19))/5

Find the zeros of 5t^2-14t+6=0

The zeros of 5t^2-14t+6=0 are t=(7+sqrt(19))/5 ,t=(7-sqrt(19))/5

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5 t^{2} - 14 t + 6 = 0

t = \frac{7 + 19}{5}, t = \frac{7 - 19}{5}

Decimal Notation

t = 2.27177 \dots, t = 0.52822 \dots

Solution steps

5 t^{2} - 14 t + 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

(- 14)^{2} - 4 \cdot 5 \cdot 6 = 219

t_{1, 2} = \frac{- ( - 14 ) \pm 2 19}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 14 ) + 2 19}{2 \cdot 5}, t_{2} = \frac{- ( - 14 ) - 2 19}{2 \cdot 5}

t = \frac{- ( - 14 ) + 2 19}{2 \cdot 5} : \frac{7 + 19}{5}

t = \frac{- ( - 14 ) - 2 19}{2 \cdot 5} : \frac{7 - 19}{5}

The solutions to the quadratic equation are:

t = \frac{7 + 19}{5}, t = \frac{7 - 19}{5}

(x - 4) (x + 5) < 0

6 t^{2} - 18 t

(2 + 2 i)^{2}

\frac{x}{8} - \frac{x}{9} = 1

x^{2} + 4 x - 6 = 0

What are the solutions to the equation 5t^2-14t+6=0 ?
The solutions to the equation 5t^2-14t+6=0 are t=(7+sqrt(19))/5 ,t=(7-sqrt(19))/5
Find the zeros of 5t^2-14t+6=0
The zeros of 5t^2-14t+6=0 are t=(7+sqrt(19))/5 ,t=(7-sqrt(19))/5