What are the solutions to the equation 0=7x+4.9x^2-8 ?

The solutions to the equation 0=7x+4.9x^2-8 are x=(-5+sqrt(105))/7 ,x=-(5+sqrt(105))/7

Find the zeros of 0=7x+4.9x^2-8

The zeros of 0=7x+4.9x^2-8 are x=(-5+sqrt(105))/7 ,x=-(5+sqrt(105))/7

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0 = 7 x + 4.9 x^{2} - 8

x = \frac{- 5 + 105}{7}, x = - \frac{5 + 105}{7}

Solution steps

0 = 7 x + 4.9 x^{2} - 8

Multiply both sides by

10

0 = 70 x + 49 x^{2} - 80

Switch sides

70 x + 49 x^{2} - 80 = 0

Write in the standard form

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

49 x^{2} + 70 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 70 \pm 7 0 ^{2} - 4 \cdot 49 ( - 80 )}{2 \cdot 49}

7 0^{2} - 4 \cdot 49 (- 80) = 14105

x_{1, 2} = \frac{- 70 \pm 14 105}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- 70 + 14 105}{2 \cdot 49}, x_{2} = \frac{- 70 - 14 105}{2 \cdot 49}

x = \frac{- 70 + 14 105}{2 \cdot 49} : \frac{- 5 + 105}{7}

x = \frac{- 70 - 14 105}{2 \cdot 49} : - \frac{5 + 105}{7}

The solutions to the quadratic equation are:

x = \frac{- 5 + 105}{7}, x = - \frac{5 + 105}{7}

7 x^{2} - 6 x - 1 = 0

5 x^{2} + 14 x - 24

x - 3 \leq 14

\frac{x ^{2}}{3} + 2 = \frac{5 x}{3}

5 \cdot 30

What are the solutions to the equation 0=7x+4.9x^2-8 ?
The solutions to the equation 0=7x+4.9x^2-8 are x=(-5+sqrt(105))/7 ,x=-(5+sqrt(105))/7
Find the zeros of 0=7x+4.9x^2-8
The zeros of 0=7x+4.9x^2-8 are x=(-5+sqrt(105))/7 ,x=-(5+sqrt(105))/7