What are the solutions to the equation 49-14x-2x^2=0 ?

The solutions to the equation 49-14x-2x^2=0 are x=-(7(1+sqrt(3)))/2 ,x=(7(sqrt(3)-1))/2

Find the zeros of 49-14x-2x^2=0

The zeros of 49-14x-2x^2=0 are x=-(7(1+sqrt(3)))/2 ,x=(7(sqrt(3)-1))/2

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49 - 14 x - 2 x^{2} = 0

x = - \frac{7 ( 1 + 3 )}{2}, x = \frac{7 ( 3 - 1 )}{2}

Decimal Notation

x = - 9.56217 \dots, x = 2.56217 \dots

Solution steps

49 - 14 x - 2 x^{2} = 0

Write in the standard form

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

- 2 x^{2} - 14 x + 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 ( - 2 ) \cdot 49}{2 ( - 2 )}

(- 14)^{2} - 4 (- 2) \cdot 49 = 143

x_{1, 2} = \frac{- ( - 14 ) \pm 14 3}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 14 3}{2 ( - 2 )}, x_{2} = \frac{- ( - 14 ) - 14 3}{2 ( - 2 )}

x = \frac{- ( - 14 ) + 14 3}{2 ( - 2 )} : - \frac{7 ( 1 + 3 )}{2}

x = \frac{- ( - 14 ) - 14 3}{2 ( - 2 )} : \frac{7 ( 3 - 1 )}{2}

The solutions to the quadratic equation are:

x = - \frac{7 ( 1 + 3 )}{2}, x = \frac{7 ( 3 - 1 )}{2}

2 d^{2} - 4 d - 5 = 0

3 q^{2} + q - 11 = - 4

(x + 24) \times (x + 2) = 420

0 = x^{2} + 5 x - 24

- 12 x - 9 = 4 x^{2}

What are the solutions to the equation 49-14x-2x^2=0 ?
The solutions to the equation 49-14x-2x^2=0 are x=-(7(1+sqrt(3)))/2 ,x=(7(sqrt(3)-1))/2
Find the zeros of 49-14x-2x^2=0
The zeros of 49-14x-2x^2=0 are x=-(7(1+sqrt(3)))/2 ,x=(7(sqrt(3)-1))/2