What are the solutions to the equation x^{(2)}+20x+28=9 ?

The solutions to the equation x^{(2)}+20x+28=9 are x=-1,x=-19

Find the zeros of x^{(2)}+20x+28=9

The zeros of x^{(2)}+20x+28=9 are x=-1,x=-19

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x^{(2)} + 20 x + 28 = 9

x = - 1, x = - 19

Solution steps

x^{(2)} + 20 x + 28 = 9

Refine

x^{2} + 20 x + 28 = 9

Move

9

to the left side

x^{2} + 20 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 19 = 18

x_{1, 2} = \frac{- 20 \pm 18}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 18}{2 \cdot 1}, x_{2} = \frac{- 20 - 18}{2 \cdot 1}

x = \frac{- 20 + 18}{2 \cdot 1} : - 1

x = \frac{- 20 - 18}{2 \cdot 1} : - 19

The solutions to the quadratic equation are:

x = - 1, x = - 19

6 - 5 x + x^{2} = 0

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

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

x^{2} - 2 x - 22 = 2 x - 1

lo g_{10} (9) = x^{2} + 3

What are the solutions to the equation x^{(2)}+20x+28=9 ?
The solutions to the equation x^{(2)}+20x+28=9 are x=-1,x=-19
Find the zeros of x^{(2)}+20x+28=9
The zeros of x^{(2)}+20x+28=9 are x=-1,x=-19