What are the solutions to the equation x(x-3)=5 ?

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

Find the zeros of x(x-3)=5

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

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x (x - 3) = 5

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

Decimal Notation

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

Solution steps

x (x - 3) = 5

Expand

x (x - 3) : x^{2} - 3 x

x^{2} - 3 x = 5

Move

5

to the left side

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^{2} - x + 42 = 0

3 x^{2} + 15 x = 0

2 x^{2} + 5 x + 6 = - 2 x

x, x^{2} + y^{2} = s^{2}

5 x^{2} - 25 x + 1 = 0

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