What are the solutions to the equation x^2-5x=4 ?

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

Find the zeros of x^2-5x=4

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

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x^{2} - 5 x = 4

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

Solution steps

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}

(3 - 5 i) (- 2 + 4 i)

y < - \frac{1}{2} (- 1) - 4

9 x > - 36

x^{2} y + x y

\frac{a - 4}{2} = 6

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