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

The solutions to the equation 0=x^2+2x-5 are x=-1+sqrt(6),x=-1-sqrt(6)

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

The zeros of 0=x^2+2x-5 are x=-1+sqrt(6),x=-1-sqrt(6)

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

x = - 1 + 6, x = - 1 - 6

Decimal Notation

x = 1.44948 \dots, x = - 3.44948 \dots

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 5) = 26

x_{1, 2} = \frac{- 2 \pm 2 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 6}{2 \cdot 1}

x = \frac{- 2 + 2 6}{2 \cdot 1} : - 1 + 6

x = \frac{- 2 - 2 6}{2 \cdot 1} : - 1 - 6

The solutions to the quadratic equation are:

x = - 1 + 6, x = - 1 - 6

- 3 (x + 2)^{2} = - 18

0.8 \cdot 5

(x + 5)^{2} = 49

\frac{3}{3} + \frac{3}{3}

lo g_{\frac{1}{64}} (16) = \frac{4}{7} x - \frac{11}{2}

What are the solutions to the equation 0=x^2+2x-5 ?
The solutions to the equation 0=x^2+2x-5 are x=-1+sqrt(6),x=-1-sqrt(6)
Find the zeros of 0=x^2+2x-5
The zeros of 0=x^2+2x-5 are x=-1+sqrt(6),x=-1-sqrt(6)