What are the solutions to the equation 17t^2-48t+28=0 ?

The solutions to the equation 17t^2-48t+28=0 are t=2,t= 14/17

Find the zeros of 17t^2-48t+28=0

The zeros of 17t^2-48t+28=0 are t=2,t= 14/17

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17 t^{2} - 48 t + 28 = 0

t = 2, t = \frac{14}{17}

Solution steps

17 t^{2} - 48 t + 28 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 17 \cdot 28}{2 \cdot 17}

(- 48)^{2} - 4 \cdot 17 \cdot 28 = 20

t_{1, 2} = \frac{- ( - 48 ) \pm 20}{2 \cdot 17}

Separate the solutions

t_{1} = \frac{- ( - 48 ) + 20}{2 \cdot 17}, t_{2} = \frac{- ( - 48 ) - 20}{2 \cdot 17}

t = \frac{- ( - 48 ) + 20}{2 \cdot 17} : 2

t = \frac{- ( - 48 ) - 20}{2 \cdot 17} : \frac{14}{17}

The solutions to the quadratic equation are:

t = 2, t = \frac{14}{17}

(2 a + 2) (a) = 60

9 p + 5 p^{2} - 15 = p^{2} + 5 p

3 (x - 7)^{2} = 30

- 4 x^{2} = 3 x^{2} - 4

3 x^{2} - 4 = 5 x

What are the solutions to the equation 17t^2-48t+28=0 ?
The solutions to the equation 17t^2-48t+28=0 are t=2,t= 14/17
Find the zeros of 17t^2-48t+28=0
The zeros of 17t^2-48t+28=0 are t=2,t= 14/17