What are the solutions to the equation 5t^2+1-7t=0 ?

The solutions to the equation 5t^2+1-7t=0 are t=(7+sqrt(29))/(10),t=(7-sqrt(29))/(10)

Find the zeros of 5t^2+1-7t=0

The zeros of 5t^2+1-7t=0 are t=(7+sqrt(29))/(10),t=(7-sqrt(29))/(10)

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5 t^{2} + 1 - 7 t = 0

t = \frac{7 + 29}{10}, t = \frac{7 - 29}{10}

Decimal Notation

t = 1.23851 \dots, t = 0.16148 \dots

Solution steps

5 t^{2} + 1 - 7 t = 0

Write in the standard form

a x^{2} + b x + c = 0

5 t^{2} - 7 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 \cdot 1 = 29

t_{1, 2} = \frac{- ( - 7 ) \pm 29}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 7 ) + 29}{2 \cdot 5}, t_{2} = \frac{- ( - 7 ) - 29}{2 \cdot 5}

t = \frac{- ( - 7 ) + 29}{2 \cdot 5} : \frac{7 + 29}{10}

t = \frac{- ( - 7 ) - 29}{2 \cdot 5} : \frac{7 - 29}{10}

The solutions to the quadratic equation are:

t = \frac{7 + 29}{10}, t = \frac{7 - 29}{10}

181.75 = - 42.8 x^{2} + 160.79 x + 145.85

2 x^{2} - 24 x + 33 = 0

5 y^{2} + 35 + 2 = 0

3 s^{2} - 9 s - 1 = 0

2 x^{2} = - 7 x - 6

What are the solutions to the equation 5t^2+1-7t=0 ?
The solutions to the equation 5t^2+1-7t=0 are t=(7+sqrt(29))/(10),t=(7-sqrt(29))/(10)
Find the zeros of 5t^2+1-7t=0
The zeros of 5t^2+1-7t=0 are t=(7+sqrt(29))/(10),t=(7-sqrt(29))/(10)