What are the solutions to the equation (x-2)-2(3x-1)=x^2 ?

The solutions to the equation (x-2)-2(3x-1)=x^2 are x=0,x=-5

Find the zeros of (x-2)-2(3x-1)=x^2

The zeros of (x-2)-2(3x-1)=x^2 are x=0,x=-5

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(x - 2) - 2 (3 x - 1) = x^{2}

x = 0, x = - 5

Solution steps

(x - 2) - 2 (3 x - 1) = x^{2}

Expand

(x - 2) - 2 (3 x - 1) : - 5 x

- 5 x = x^{2}

Switch sides

x^{2} = - 5 x

Move

5 x

to the left side

x^{2} + 5 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

x_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 5}{2 \cdot 1}, x_{2} = \frac{- 5 - 5}{2 \cdot 1}

x = \frac{- 5 + 5}{2 \cdot 1} : 0

x = \frac{- 5 - 5}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 0, x = - 5

z (2 z - 3) = 14

101 x^{2} + 52 x + 16 = 0

4 x^{2} + 7 x = 3

x^{2} - 14 x + 11 = 11

(2 x + 1)^{2} + 5 x = 10

What are the solutions to the equation (x-2)-2(3x-1)=x^2 ?
The solutions to the equation (x-2)-2(3x-1)=x^2 are x=0,x=-5
Find the zeros of (x-2)-2(3x-1)=x^2
The zeros of (x-2)-2(3x-1)=x^2 are x=0,x=-5