What are the solutions to the equation (x^2-2x)+(6x+4)=0 ?

The solutions to the equation (x^2-2x)+(6x+4)=0 are x=-2

Find the zeros of (x^2-2x)+(6x+4)=0

The zeros of (x^2-2x)+(6x+4)=0 are x=-2

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(x^{2} - 2 x) + (6 x + 4) = 0

x = - 2

Solution steps

(x^{2} - 2 x) + (6 x + 4) = 0

Refine

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

Solve with the quadratic formula

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

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

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

x = \frac{- 4}{2 \cdot 1}

\frac{- 4}{2 \cdot 1} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

(9 x + 4)^{2} = 16

3 x (x + 5) = - 8

64 x^{2} - 121 = 0

y, (z - 4)^{2} = 4 y^{2}

20000 - 200 Q + Q^{2} = 20000 - 400 Q + 3 Q^{2}

What are the solutions to the equation (x^2-2x)+(6x+4)=0 ?
The solutions to the equation (x^2-2x)+(6x+4)=0 are x=-2
Find the zeros of (x^2-2x)+(6x+4)=0
The zeros of (x^2-2x)+(6x+4)=0 are x=-2