What are the solutions to the equation 104=124x-14x^2 ?

The solutions to the equation 104=124x-14x^2 are x=(31-sqrt(597))/7 ,x=(31+sqrt(597))/7

Find the zeros of 104=124x-14x^2

The zeros of 104=124x-14x^2 are x=(31-sqrt(597))/7 ,x=(31+sqrt(597))/7

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104 = 124 x - 14 x^{2}

x = \frac{31 - 597}{7}, x = \frac{31 + 597}{7}

Decimal

x = 0.93805 \dots, x = 7.91908 \dots

Solution steps

104 = 124 x - 14 x^{2}

Switch sides

124 x - 14 x^{2} = 104

Move

104

to the left side

124 x - 14 x^{2} - 104 = 0

Write in the standard form

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

- 14 x^{2} + 124 x - 104 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 124 \pm 12 4 ^{2} - 4 ( - 14 ) ( - 104 )}{2 ( - 14 )}

12 4^{2} - 4 (- 14) (- 104) = 4597

x_{1, 2} = \frac{- 124 \pm 4 597}{2 ( - 14 )}

Separate the solutions

x_{1} = \frac{- 124 + 4 597}{2 ( - 14 )}, x_{2} = \frac{- 124 - 4 597}{2 ( - 14 )}

x = \frac{- 124 + 4 597}{2 ( - 14 )} : \frac{31 - 597}{7}

x = \frac{- 124 - 4 597}{2 ( - 14 )} : \frac{31 + 597}{7}

The solutions to the quadratic equation are:

x = \frac{31 - 597}{7}, x = \frac{31 + 597}{7}

2 c^{2} - 12 c + 7 = - 4 c

x^{2} - 7

4 y^{2} + 3 y - 1 = 0

x, 5 a x^{2} y - 5 x^{2} + 6 y^{3} = 100

(3 x + 1) (3 x - 1) = 5 x + 3

What are the solutions to the equation 104=124x-14x^2 ?
The solutions to the equation 104=124x-14x^2 are x=(31-sqrt(597))/7 ,x=(31+sqrt(597))/7
Find the zeros of 104=124x-14x^2
The zeros of 104=124x-14x^2 are x=(31-sqrt(597))/7 ,x=(31+sqrt(597))/7