What are the solutions to the equation 4.9x^2-29.4x=12 ?

The solutions to the equation 4.9x^2-29.4x=12 are x=(21+sqrt(561))/7 ,x=(21-sqrt(561))/7

Find the zeros of 4.9x^2-29.4x=12

The zeros of 4.9x^2-29.4x=12 are x=(21+sqrt(561))/7 ,x=(21-sqrt(561))/7

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4.9 x^{2} - 29.4 x = 12

x = \frac{21 + 561}{7}, x = \frac{21 - 561}{7}

Solution steps

4.9 x^{2} - 29.4 x = 12

Multiply both sides by

10

49 x^{2} - 294 x = 120

Move

120

to the left side

49 x^{2} - 294 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 294 ) \pm ( - 294 ) ^{2} - 4 \cdot 49 ( - 120 )}{2 \cdot 49}

(- 294)^{2} - 4 \cdot 49 (- 120) = 14561

x_{1, 2} = \frac{- ( - 294 ) \pm 14 561}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- ( - 294 ) + 14 561}{2 \cdot 49}, x_{2} = \frac{- ( - 294 ) - 14 561}{2 \cdot 49}

x = \frac{- ( - 294 ) + 14 561}{2 \cdot 49} : \frac{21 + 561}{7}

x = \frac{- ( - 294 ) - 14 561}{2 \cdot 49} : \frac{21 - 561}{7}

The solutions to the quadratic equation are:

x = \frac{21 + 561}{7}, x = \frac{21 - 561}{7}

10 x^{2} + 40 x + 14 x + 56

z^{2} - 121

x y + x - y - 1

y^{2} - 25

2 r^{2} - 11 r + 5

What are the solutions to the equation 4.9x^2-29.4x=12 ?
The solutions to the equation 4.9x^2-29.4x=12 are x=(21+sqrt(561))/7 ,x=(21-sqrt(561))/7
Find the zeros of 4.9x^2-29.4x=12
The zeros of 4.9x^2-29.4x=12 are x=(21+sqrt(561))/7 ,x=(21-sqrt(561))/7