What are the solutions to the equation 13-13x=(4x+3)(x-1) ?

The solutions to the equation 13-13x=(4x+3)(x-1) are x=1,x=-4

Find the zeros of 13-13x=(4x+3)(x-1)

The zeros of 13-13x=(4x+3)(x-1) are x=1,x=-4

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13 - 13 x = (4 x + 3) (x - 1)

x = 1, x = - 4

Solution steps

13 - 13 x = (4 x + 3) (x - 1)

Expand

(4 x + 3) (x - 1) : 4 x^{2} - x - 3

13 - 13 x = 4 x^{2} - x - 3

Switch sides

4 x^{2} - x - 3 = 13 - 13 x

Move

13 x

to the left side

4 x^{2} + 12 x - 3 = 13

Move

13

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

5 x^{2} + 10 x + 78 = 8

0.5 x^{2} + 0.875 x - 0.0525 = 0

x, x^{2} + 4 x + 16 = 0

x (2 x + 4) + 5 x = - 7

9 x^{2} + 18 x + 82 = 0

What are the solutions to the equation 13-13x=(4x+3)(x-1) ?
The solutions to the equation 13-13x=(4x+3)(x-1) are x=1,x=-4
Find the zeros of 13-13x=(4x+3)(x-1)
The zeros of 13-13x=(4x+3)(x-1) are x=1,x=-4