What are the solutions to the equation x^2+445x-2225=0 ?

The solutions to the equation x^2+445x-2225=0 are x=(-445+5sqrt(8277))/2 ,x=(-445-5sqrt(8277))/2

Find the zeros of x^2+445x-2225=0

The zeros of x^2+445x-2225=0 are x=(-445+5sqrt(8277))/2 ,x=(-445-5sqrt(8277))/2

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x^{2} + 445 x - 2225 = 0

x = \frac{- 445 + 5 8277}{2}, x = \frac{- 445 - 5 8277}{2}

Decimal Notation

x = 4.94504 \dots, x = - 449.94504 \dots

Solution steps

x^{2} + 445 x - 2225 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 445 \pm 44 5 ^{2} - 4 \cdot 1 \cdot ( - 2225 )}{2 \cdot 1}

44 5^{2} - 4 \cdot 1 \cdot (- 2225) = 58277

x_{1, 2} = \frac{- 445 \pm 5 8277}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 445 + 5 8277}{2 \cdot 1}, x_{2} = \frac{- 445 - 5 8277}{2 \cdot 1}

x = \frac{- 445 + 5 8277}{2 \cdot 1} : \frac{- 445 + 5 8277}{2}

x = \frac{- 445 - 5 8277}{2 \cdot 1} : \frac{- 445 - 5 8277}{2}

The solutions to the quadratic equation are:

x = \frac{- 445 + 5 8277}{2}, x = \frac{- 445 - 5 8277}{2}

x, 4 x^{2} + 4 y^{2} = z^{2}

- 3 x^{2} - 3 x + 10

a^{2} = 3^{2} + 4^{2}

m (m - 1) + 11 m + 5 = 0

5 r^{2} + 27 r - 18 = 0

What are the solutions to the equation x^2+445x-2225=0 ?
The solutions to the equation x^2+445x-2225=0 are x=(-445+5sqrt(8277))/2 ,x=(-445-5sqrt(8277))/2
Find the zeros of x^2+445x-2225=0
The zeros of x^2+445x-2225=0 are x=(-445+5sqrt(8277))/2 ,x=(-445-5sqrt(8277))/2