What are the solutions to the equation x(x-5)=4 ?

The solutions to the equation x(x-5)=4 are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2

Find the zeros of x(x-5)=4

The zeros of x(x-5)=4 are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2

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x (x - 5) = 4

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

Decimal Notation

x = 5.70156 \dots, x = - 0.70156 \dots

Solution steps

x (x - 5) = 4

Expand

x (x - 5) : x^{2} - 5 x

x^{2} - 5 x = 4

Move

4

to the left side

x^{2} - 5 x - 4 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 1 \cdot (- 4) = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 1}

x = \frac{- ( - 5 ) + 41}{2 \cdot 1} : \frac{5 + 41}{2}

x = \frac{- ( - 5 ) - 41}{2 \cdot 1} : \frac{5 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

x (x - 5) = 2

25 y^{2} + 15 y + 21 = 0

7 x (1 - x) = 5 (x - 2)

x^{2} - x - 2 = 0

x, x^{2} + 3 x - 2 = 0

What are the solutions to the equation x(x-5)=4 ?
The solutions to the equation x(x-5)=4 are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2
Find the zeros of x(x-5)=4
The zeros of x(x-5)=4 are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2