What are the solutions to the equation x^2+5x+1^2=4 ?

The solutions to the equation x^2+5x+1^2=4 are x=(-5+sqrt(37))/2 ,x=(-5-sqrt(37))/2

Find the zeros of x^2+5x+1^2=4

The zeros of x^2+5x+1^2=4 are x=(-5+sqrt(37))/2 ,x=(-5-sqrt(37))/2

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x^{2} + 5 x + 1^{2} = 4

x = \frac{- 5 + 37}{2}, x = \frac{- 5 - 37}{2}

Decimal Notation

x = 0.54138 \dots, x = - 5.54138 \dots

Solution steps

x^{2} + 5 x + 1^{2} = 4

Expand

x^{2} + 5 x + 1^{2} : x^{2} + 5 x + 1

x^{2} + 5 x + 1 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 3) = 37

x_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 37}{2 \cdot 1}, x_{2} = \frac{- 5 - 37}{2 \cdot 1}

x = \frac{- 5 + 37}{2 \cdot 1} : \frac{- 5 + 37}{2}

x = \frac{- 5 - 37}{2 \cdot 1} : \frac{- 5 - 37}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 37}{2}, x = \frac{- 5 - 37}{2}

y^{2} + 10 y - 26 = 0

- 3 x^{2} - 51 x - 216 = 0

6 (x + 12) = x^{2}

m^{2} + 6 m + 18 = 0

x^{2} + 7 x + \frac{49}{4} = 4

What are the solutions to the equation x^2+5x+1^2=4 ?
The solutions to the equation x^2+5x+1^2=4 are x=(-5+sqrt(37))/2 ,x=(-5-sqrt(37))/2
Find the zeros of x^2+5x+1^2=4
The zeros of x^2+5x+1^2=4 are x=(-5+sqrt(37))/2 ,x=(-5-sqrt(37))/2