What are the solutions to the equation x^2+3x=1 ?

The solutions to the equation x^2+3x=1 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2

Find the zeros of x^2+3x=1

The zeros of x^2+3x=1 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2

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x^{2} + 3 x = 1

x = \frac{- 3 + 13}{2}, x = \frac{- 3 - 13}{2}

Decimal Notation

x = 0.30277 \dots, x = - 3.30277 \dots

Solution steps

x^{2} + 3 x = 1

Move

1

to the left side

x^{2} + 3 x - 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

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

Separate the solutions

x_{1} = \frac{- 3 + 13}{2 \cdot 1}, x_{2} = \frac{- 3 - 13}{2 \cdot 1}

x = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

x = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 13}{2}, x = \frac{- 3 - 13}{2}

5 (8) - 2 y = 30

17 a x - 17 m x + 2 a y - 2 m y + 9 a z - 9 m z

(\frac{5}{3})^{1}

∣ 2 x + 7 ∣ \geq 19

3 (v - 2) + 5 = 2 (3 v - 5)

What are the solutions to the equation x^2+3x=1 ?
The solutions to the equation x^2+3x=1 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2
Find the zeros of x^2+3x=1
The zeros of x^2+3x=1 are x=(-3+sqrt(13))/2 ,x=(-3-sqrt(13))/2