What are the solutions to the equation x^2+12x=7x+1 ?

The solutions to the equation x^2+12x=7x+1 are x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2

Find the zeros of x^2+12x=7x+1

The zeros of x^2+12x=7x+1 are x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2

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x^{2} + 12 x = 7 x + 1

x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

Decimal Notation

x = 0.19258 \dots, x = - 5.19258 \dots

Solution steps

x^{2} + 12 x = 7 x + 1

Move

1

to the left side

x^{2} + 12 x - 1 = 7 x

Move

7 x

to the left side

x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 1) = 29

x_{1, 2} = \frac{- 5 \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 29}{2 \cdot 1}, x_{2} = \frac{- 5 - 29}{2 \cdot 1}

x = \frac{- 5 + 29}{2 \cdot 1} : \frac{- 5 + 29}{2}

x = \frac{- 5 - 29}{2 \cdot 1} : \frac{- 5 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

2 x^{2} - 8 x y - 42 y^{2}

\frac{1}{3 ( 0 ) ^{\frac{2}{3}}}

\frac{x - 8}{x + 3} > 0

\frac{1}{2} (1 - 0.88)

\frac{- 1}{3} + \frac{- 1}{3}

What are the solutions to the equation x^2+12x=7x+1 ?
The solutions to the equation x^2+12x=7x+1 are x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2
Find the zeros of x^2+12x=7x+1
The zeros of x^2+12x=7x+1 are x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2