What are the solutions to the equation -4x^2-42x-98=-2x-2 ?

The solutions to the equation -4x^2-42x-98=-2x-2 are x=-6,x=-4

Find the zeros of -4x^2-42x-98=-2x-2

The zeros of -4x^2-42x-98=-2x-2 are x=-6,x=-4

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- 4 x^{2} - 42 x - 98 = - 2 x - 2

x = - 6, x = - 4

Solution steps

- 4 x^{2} - 42 x - 98 = - 2 x - 2

Move

2

to the left side

- 4 x^{2} - 42 x - 96 = - 2 x

Move

2 x

to the left side

- 4 x^{2} - 40 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 4 ) ( - 96 )}{2 ( - 4 )}

(- 40)^{2} - 4 (- 4) (- 96) = 8

x_{1, 2} = \frac{- ( - 40 ) \pm 8}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 8}{2 ( - 4 )}, x_{2} = \frac{- ( - 40 ) - 8}{2 ( - 4 )}

x = \frac{- ( - 40 ) + 8}{2 ( - 4 )} : - 6

x = \frac{- ( - 40 ) - 8}{2 ( - 4 )} : - 4

The solutions to the quadratic equation are:

x = - 6, x = - 4

\frac{2 x}{5} - \frac{1}{3} = \frac{137}{3}

- \frac{1}{5} - (- \frac{2}{9})

ln (5 - 7 x) \leq 1

36 \cdot 5

5 y^{2} + 16 y + 3

What are the solutions to the equation -4x^2-42x-98=-2x-2 ?
The solutions to the equation -4x^2-42x-98=-2x-2 are x=-6,x=-4
Find the zeros of -4x^2-42x-98=-2x-2
The zeros of -4x^2-42x-98=-2x-2 are x=-6,x=-4