What are the solutions to the equation x^2+3x=25 ?

The solutions to the equation x^2+3x=25 are x=(-3+sqrt(109))/2 ,x=(-3-sqrt(109))/2

Find the zeros of x^2+3x=25

The zeros of x^2+3x=25 are x=(-3+sqrt(109))/2 ,x=(-3-sqrt(109))/2

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x^{2} + 3 x = 25

x = \frac{- 3 + 109}{2}, x = \frac{- 3 - 109}{2}

Decimal Notation

x = 3.72015 \dots, x = - 6.72015 \dots

Solution steps

x^{2} + 3 x = 25

Move

25

to the left side

x^{2} + 3 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 25) = 109

x_{1, 2} = \frac{- 3 \pm 109}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 109}{2 \cdot 1}, x_{2} = \frac{- 3 - 109}{2 \cdot 1}

x = \frac{- 3 + 109}{2 \cdot 1} : \frac{- 3 + 109}{2}

x = \frac{- 3 - 109}{2 \cdot 1} : \frac{- 3 - 109}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 109}{2}, x = \frac{- 3 - 109}{2}

24 3^{- \frac{5}{6}} \cdot 3^{- \frac{5}{6}}

- 25 - 7 x = 6 (2 x - 1)

x^{4} - 26 x^{2} + 25 \geq 0

\frac{5}{2} - 3

3 x^{2} - 17 x > 28

What are the solutions to the equation x^2+3x=25 ?
The solutions to the equation x^2+3x=25 are x=(-3+sqrt(109))/2 ,x=(-3-sqrt(109))/2
Find the zeros of x^2+3x=25
The zeros of x^2+3x=25 are x=(-3+sqrt(109))/2 ,x=(-3-sqrt(109))/2