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

The solutions to the equation x^2+3x=2 are x=(-3+sqrt(17))/2 ,x=(-3-sqrt(17))/2

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

The zeros of x^2+3x=2 are x=(-3+sqrt(17))/2 ,x=(-3-sqrt(17))/2

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x^{2} + 3 x = 2

x = \frac{- 3 + 17}{2}, x = \frac{- 3 - 17}{2}

Decimal Notation

x = 0.56155 \dots, x = - 3.56155 \dots

Solution steps

x^{2} + 3 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 2) = 17

x_{1, 2} = \frac{- 3 \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 17}{2 \cdot 1}, x_{2} = \frac{- 3 - 17}{2 \cdot 1}

x = \frac{- 3 + 17}{2 \cdot 1} : \frac{- 3 + 17}{2}

x = \frac{- 3 - 17}{2 \cdot 1} : \frac{- 3 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 17}{2}, x = \frac{- 3 - 17}{2}

7 x^{2} - 12 x - 4 = 0

3 p + 13 = 37

5 x^{2} + 8 x + 15 = 0

(4 + i) (3 + 2 i) (1 - i)

\frac{3}{2} + \frac{1}{3} - \frac{1}{4}

What are the solutions to the equation x^2+3x=2 ?
The solutions to the equation x^2+3x=2 are x=(-3+sqrt(17))/2 ,x=(-3-sqrt(17))/2
Find the zeros of x^2+3x=2
The zeros of x^2+3x=2 are x=(-3+sqrt(17))/2 ,x=(-3-sqrt(17))/2