What are the solutions to the equation x^2+12x+48=-x+6 ?

The solutions to the equation x^2+12x+48=-x+6 are x=-6,x=-7

Find the zeros of x^2+12x+48=-x+6

The zeros of x^2+12x+48=-x+6 are x=-6,x=-7

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x^{2} + 12 x + 48 = - x + 6

x = - 6, x = - 7

Solution steps

x^{2} + 12 x + 48 = - x + 6

Move

6

to the left side

x^{2} + 12 x + 42 = - x

Move

x

to the left side

x^{2} + 13 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 42 = 1

x_{1, 2} = \frac{- 13 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 1}{2 \cdot 1}, x_{2} = \frac{- 13 - 1}{2 \cdot 1}

x = \frac{- 13 + 1}{2 \cdot 1} : - 6

x = \frac{- 13 - 1}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = - 6, x = - 7

2 (- 2) + 1

4 x^{2} - 4 x = - 7

∣ x - 4 ∣ = 0

7 (2 y - 8) = 14

2 a^{2} = - 6 + 8 a

What are the solutions to the equation x^2+12x+48=-x+6 ?
The solutions to the equation x^2+12x+48=-x+6 are x=-6,x=-7
Find the zeros of x^2+12x+48=-x+6
The zeros of x^2+12x+48=-x+6 are x=-6,x=-7