What are the solutions to the equation 0=x^2-3x-5 ?

The solutions to the equation 0=x^2-3x-5 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

Find the zeros of 0=x^2-3x-5

The zeros of 0=x^2-3x-5 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

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0 = x^{2} - 3 x - 5

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

Decimal Notation

x = 4.19258 \dots, x = - 1.19258 \dots

Solution steps

0 = x^{2} - 3 x - 5

Switch sides

x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 5) = 29

x_{1, 2} = \frac{- ( - 3 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 29}{2 \cdot 1}

x = \frac{- ( - 3 ) + 29}{2 \cdot 1} : \frac{3 + 29}{2}

x = \frac{- ( - 3 ) - 29}{2 \cdot 1} : \frac{3 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

x, x^{2} + 2 y x + 1 = 0

x^{2} + 18 x = - 17

8 x^{2} - 1 - x = - x + 5

x^{2} + 18 x = - 45

(x - 5)^{2} + 17 = 9

What are the solutions to the equation 0=x^2-3x-5 ?
The solutions to the equation 0=x^2-3x-5 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2
Find the zeros of 0=x^2-3x-5
The zeros of 0=x^2-3x-5 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2