What are the solutions to the equation x(x+3)=5 ?

The solutions to the equation x(x+3)=5 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2

Find the zeros of x(x+3)=5

The zeros of x(x+3)=5 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2

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x (x + 3) = 5

x = \frac{- 3 + 29}{2}, x = \frac{- 3 - 29}{2}

Decimal Notation

x = 1.19258 \dots, x = - 4.19258 \dots

Solution steps

x (x + 3) = 5

Expand

x (x + 3) : x^{2} + 3 x

x^{2} + 3 x = 5

Move

5

to the left side

x^{2} + 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 5) = 29

x_{1, 2} = \frac{- 3 \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 29}{2 \cdot 1}, x_{2} = \frac{- 3 - 29}{2 \cdot 1}

x = \frac{- 3 + 29}{2 \cdot 1} : \frac{- 3 + 29}{2}

x = \frac{- 3 - 29}{2 \cdot 1} : \frac{- 3 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 29}{2}, x = \frac{- 3 - 29}{2}

- 7 x^{5} - 56 x^{4}

1 x = 2

\frac{15}{37.81 + 78 + 22.85 + 13.15}

(6 x^{2} 4 y^{- 5}) (\frac{1}{3} x^{- 3} y^{2})

12 y + 3 y = 30

What are the solutions to the equation x(x+3)=5 ?
The solutions to the equation x(x+3)=5 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2
Find the zeros of x(x+3)=5
The zeros of x(x+3)=5 are x=(-3+sqrt(29))/2 ,x=(-3-sqrt(29))/2