What are the solutions to the equation 56(x+1)+56x=-15x(x+1) ?

The solutions to the equation 56(x+1)+56x=-15x(x+1) are x=-8,x=-7/15

Find the zeros of 56(x+1)+56x=-15x(x+1)

The zeros of 56(x+1)+56x=-15x(x+1) are x=-8,x=-7/15

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56 (x + 1) + 56 x = - 15 x (x + 1)

x = - 8, x = - \frac{7}{15}

Decimal Notation

x = - 8, x = - 0.46666 \dots

Solution steps

56 (x + 1) + 56 x = - 15 x (x + 1)

Expand

56 (x + 1) + 56 x : 112 x + 56

Expand

- 15 x (x + 1) : - 15 x^{2} - 15 x

112 x + 56 = - 15 x^{2} - 15 x

Switch sides

- 15 x^{2} - 15 x = 112 x + 56

Move

56

to the left side

- 15 x^{2} - 15 x - 56 = 112 x

Move

112 x

to the left side

- 15 x^{2} - 127 x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 127 ) \pm ( - 127 ) ^{2} - 4 ( - 15 ) ( - 56 )}{2 ( - 15 )}

(- 127)^{2} - 4 (- 15) (- 56) = 113

x_{1, 2} = \frac{- ( - 127 ) \pm 113}{2 ( - 15 )}

Separate the solutions

x_{1} = \frac{- ( - 127 ) + 113}{2 ( - 15 )}, x_{2} = \frac{- ( - 127 ) - 113}{2 ( - 15 )}

x = \frac{- ( - 127 ) + 113}{2 ( - 15 )} : - 8

x = \frac{- ( - 127 ) - 113}{2 ( - 15 )} : - \frac{7}{15}

The solutions to the quadratic equation are:

x = - 8, x = - \frac{7}{15}

x^{2} - 2

x^{2} - 10 x = - 40

x^{2} - 10 x = - 23

6 t^{2} - 18 t = 0

3 x^{2} = - 2 x + 7

What are the solutions to the equation 56(x+1)+56x=-15x(x+1) ?
The solutions to the equation 56(x+1)+56x=-15x(x+1) are x=-8,x=-7/15
Find the zeros of 56(x+1)+56x=-15x(x+1)
The zeros of 56(x+1)+56x=-15x(x+1) are x=-8,x=-7/15