What are the solutions to the equation x^2+7x+5=0 ?

The solutions to the equation x^2+7x+5=0 are x=(-7+sqrt(29))/2 ,x=(-7-sqrt(29))/2

Find the zeros of x^2+7x+5=0

The zeros of x^2+7x+5=0 are x=(-7+sqrt(29))/2 ,x=(-7-sqrt(29))/2

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x^{2} + 7 x + 5 = 0

x = \frac{- 7 + 29}{2}, x = \frac{- 7 - 29}{2}

Decimal Notation

x = - 0.80741 \dots, x = - 6.19258 \dots

Solution steps

x^{2} + 7 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 5 = 29

x_{1, 2} = \frac{- 7 \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 29}{2 \cdot 1}, x_{2} = \frac{- 7 - 29}{2 \cdot 1}

x = \frac{- 7 + 29}{2 \cdot 1} : \frac{- 7 + 29}{2}

x = \frac{- 7 - 29}{2 \cdot 1} : \frac{- 7 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 29}{2}, x = \frac{- 7 - 29}{2}

12 n^{3} - 15 n^{2} - 8 n + 10

(a^{\frac{2}{5}})^{- \frac{3}{4}}

m^{2} - 4 mn - 21 n^{2}

(\frac{1}{243})^{\frac{1}{5}}

\frac{- 1}{2} + 2

What are the solutions to the equation x^2+7x+5=0 ?
The solutions to the equation x^2+7x+5=0 are x=(-7+sqrt(29))/2 ,x=(-7-sqrt(29))/2
Find the zeros of x^2+7x+5=0
The zeros of x^2+7x+5=0 are x=(-7+sqrt(29))/2 ,x=(-7-sqrt(29))/2