What are the solutions to the equation x(13+x)+6=42 ?

The solutions to the equation x(13+x)+6=42 are x=(-13+sqrt(313))/2 ,x=(-13-sqrt(313))/2

Find the zeros of x(13+x)+6=42

The zeros of x(13+x)+6=42 are x=(-13+sqrt(313))/2 ,x=(-13-sqrt(313))/2

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x (13 + x) + 6 = 42

x = \frac{- 13 + 313}{2}, x = \frac{- 13 - 313}{2}

Decimal Notation

x = 2.34590 \dots, x = - 15.34590 \dots

Solution steps

x (13 + x) + 6 = 42

Expand

x (13 + x) + 6 : 13 x + x^{2} + 6

13 x + x^{2} + 6 = 42

Move

42

to the left side

x^{2} + 13 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot (- 36) = 313

x_{1, 2} = \frac{- 13 \pm 313}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 313}{2 \cdot 1}, x_{2} = \frac{- 13 - 313}{2 \cdot 1}

x = \frac{- 13 + 313}{2 \cdot 1} : \frac{- 13 + 313}{2}

x = \frac{- 13 - 313}{2 \cdot 1} : \frac{- 13 - 313}{2}

The solutions to the quadratic equation are:

x = \frac{- 13 + 313}{2}, x = \frac{- 13 - 313}{2}

x^{2} + 4 x + 1 = 0

6 x = x^{2} - 27

3 x^{2} - 12 x + 27 = 0

a^{2} - 7 a

4 x^{2} - 8 x = 52

What are the solutions to the equation x(13+x)+6=42 ?
The solutions to the equation x(13+x)+6=42 are x=(-13+sqrt(313))/2 ,x=(-13-sqrt(313))/2
Find the zeros of x(13+x)+6=42
The zeros of x(13+x)+6=42 are x=(-13+sqrt(313))/2 ,x=(-13-sqrt(313))/2