What are the solutions to the equation 104=u^2-5u ?

The solutions to the equation 104=u^2-5u are u=13,u=-8

Find the zeros of 104=u^2-5u

The zeros of 104=u^2-5u are u=13,u=-8

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104 = u^{2} - 5 u

u = 13, u = - 8

Solution steps

104 = u^{2} - 5 u

Switch sides

u^{2} - 5 u = 104

Move

104

to the left side

u^{2} - 5 u - 104 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 104 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 104) = 21

u_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

u = \frac{- ( - 5 ) + 21}{2 \cdot 1} : 13

u = \frac{- ( - 5 ) - 21}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

u = 13, u = - 8

(x - 2)^{2} + (x + 2)^{2} = 0

- x^{2} - 2 x + 19 = 0

x^{2} + 4 x + 64 = 0

(7 - x) (- 3 - x) + 26 = 0

3 x^{2} - 6 x + 5 = 0

What are the solutions to the equation 104=u^2-5u ?
The solutions to the equation 104=u^2-5u are u=13,u=-8
Find the zeros of 104=u^2-5u
The zeros of 104=u^2-5u are u=13,u=-8