What are the solutions to the equation x^2+20x+40=0 ?

The solutions to the equation x^2+20x+40=0 are x=2(sqrt(15)-5),x=-2(5+sqrt(15))

Find the zeros of x^2+20x+40=0

The zeros of x^2+20x+40=0 are x=2(sqrt(15)-5),x=-2(5+sqrt(15))

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x^{2} + 20 x + 40 = 0

x = 2 (15 - 5), x = - 2 (5 + 15)

Decimal Notation

x = - 2.25403 \dots, x = - 17.74596 \dots

Solution steps

x^{2} + 20 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 40 = 415

x_{1, 2} = \frac{- 20 \pm 4 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 4 15}{2 \cdot 1}, x_{2} = \frac{- 20 - 4 15}{2 \cdot 1}

x = \frac{- 20 + 4 15}{2 \cdot 1} : 2 (15 - 5)

x = \frac{- 20 - 4 15}{2 \cdot 1} : - 2 (5 + 15)

The solutions to the quadratic equation are:

x = 2 (15 - 5), x = - 2 (5 + 15)

(3 x^{5} - \frac{1}{9} y^{3})^{4}

10 y^{2} + 11 y + 3

j^{2} + 24 j - 25

\frac{4 x ^{5} y ^{7}}{8 x ^{2} y}

f = 1^{2} - 3

What are the solutions to the equation x^2+20x+40=0 ?
The solutions to the equation x^2+20x+40=0 are x=2(sqrt(15)-5),x=-2(5+sqrt(15))
Find the zeros of x^2+20x+40=0
The zeros of x^2+20x+40=0 are x=2(sqrt(15)-5),x=-2(5+sqrt(15))