What are the solutions to the equation 4.9t^2-78.4t+73.5=0 ?

The solutions to the equation 4.9t^2-78.4t+73.5=0 are t=15,t=1

Find the zeros of 4.9t^2-78.4t+73.5=0

The zeros of 4.9t^2-78.4t+73.5=0 are t=15,t=1

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4.9 t^{2} - 78.4 t + 73.5 = 0

t = 15, t = 1

Solution steps

4.9 t^{2} - 78.4 t + 73.5 = 0

Multiply both sides by

10

49 t^{2} - 784 t + 735 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 784 ) \pm ( - 784 ) ^{2} - 4 \cdot 49 \cdot 735}{2 \cdot 49}

(- 784)^{2} - 4 \cdot 49 \cdot 735 = 686

t_{1, 2} = \frac{- ( - 784 ) \pm 686}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- ( - 784 ) + 686}{2 \cdot 49}, t_{2} = \frac{- ( - 784 ) - 686}{2 \cdot 49}

t = \frac{- ( - 784 ) + 686}{2 \cdot 49} : 15

t = \frac{- ( - 784 ) - 686}{2 \cdot 49} : 1

The solutions to the quadratic equation are:

t = 15, t = 1

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

(1 - x) (- 1 - x) + 2 = 0

(3 z - 4) (3 z + 4) = 0

(x 2 + 2 x - 2) \cdot (m x - n) = 5 x 3 + 8 x 2 - 14 x + 4

x^{2} + 62.5 = 0

What are the solutions to the equation 4.9t^2-78.4t+73.5=0 ?
The solutions to the equation 4.9t^2-78.4t+73.5=0 are t=15,t=1
Find the zeros of 4.9t^2-78.4t+73.5=0
The zeros of 4.9t^2-78.4t+73.5=0 are t=15,t=1