What are the solutions to the equation 7-7x=(4x+9)(x-1) ?

The solutions to the equation 7-7x=(4x+9)(x-1) are x=1,x=-4

Find the zeros of 7-7x=(4x+9)(x-1)

The zeros of 7-7x=(4x+9)(x-1) are x=1,x=-4

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7 - 7 x = (4 x + 9) (x - 1)

x = 1, x = - 4

Solution steps

7 - 7 x = (4 x + 9) (x - 1)

Expand

(4 x + 9) (x - 1) : 4 x^{2} + 5 x - 9

7 - 7 x = 4 x^{2} + 5 x - 9

Switch sides

4 x^{2} + 5 x - 9 = 7 - 7 x

Move

7 x

to the left side

4 x^{2} + 12 x - 9 = 7

Move

7

to the left side

4 x^{2} + 12 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 16 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 16) = 20

x_{1, 2} = \frac{- 12 \pm 20}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 20}{2 \cdot 4}, x_{2} = \frac{- 12 - 20}{2 \cdot 4}

x = \frac{- 12 + 20}{2 \cdot 4} : 1

x = \frac{- 12 - 20}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

\frac{12 x y ^{7}}{6 x ^{6} y ^{5}}

- 3.28 - (- 4.4) + (- 9.7)

2 (1) + 1

3 x - 4 (2 x - 5) = 5

\frac{3}{2} \cdot 12

What are the solutions to the equation 7-7x=(4x+9)(x-1) ?
The solutions to the equation 7-7x=(4x+9)(x-1) are x=1,x=-4
Find the zeros of 7-7x=(4x+9)(x-1)
The zeros of 7-7x=(4x+9)(x-1) are x=1,x=-4