What are the solutions to the equation 2=x^2+5x ?

The solutions to the equation 2=x^2+5x are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2

Find the zeros of 2=x^2+5x

The zeros of 2=x^2+5x are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2

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2 = x^{2} + 5 x

x = \frac{- 5 + 33}{2}, x = \frac{- 5 - 33}{2}

Decimal Notation

x = 0.37228 \dots, x = - 5.37228 \dots

Solution steps

2 = x^{2} + 5 x

Switch sides

x^{2} + 5 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 2) = 33

x_{1, 2} = \frac{- 5 \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 33}{2 \cdot 1}, x_{2} = \frac{- 5 - 33}{2 \cdot 1}

x = \frac{- 5 + 33}{2 \cdot 1} : \frac{- 5 + 33}{2}

x = \frac{- 5 - 33}{2 \cdot 1} : \frac{- 5 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 33}{2}, x = \frac{- 5 - 33}{2}

27 x^{3} + 512 y^{3}

6.5 = 0.5 \cdot 1200 \cdot x^{2}

50 - [3 (2 x - 3) - 2 (2 x - 1)] = 23

\frac{4}{7} + \frac{1}{7}

y^{- \frac{5}{4}}

What are the solutions to the equation 2=x^2+5x ?
The solutions to the equation 2=x^2+5x are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2
Find the zeros of 2=x^2+5x
The zeros of 2=x^2+5x are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2