What are the solutions to the equation x^2+5x+2=4 ?

The solutions to the equation x^2+5x+2=4 are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2

Find the zeros of x^2+5x+2=4

The zeros of x^2+5x+2=4 are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2

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x^{2} + 5 x + 2 = 4

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

Solution steps

x^{2} + 5 x + 2 = 4

Move

4

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}

(2 x + 3) (3 x - 7) = 0

3 (x + 4)^{2} = 8

- 4 + 16 x - 15 x^{2} = 0

4 x^{2} - 20 x + 29 = 0

(x + 5)^{2} + 5 (2 x + 3) = 2 (x + 4)^{2}

What are the solutions to the equation x^2+5x+2=4 ?
The solutions to the equation x^2+5x+2=4 are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2
Find the zeros of x^2+5x+2=4
The zeros of x^2+5x+2=4 are x=(-5+sqrt(33))/2 ,x=(-5-sqrt(33))/2