What are the solutions to the equation u^2-4u=12 ?

The solutions to the equation u^2-4u=12 are u=6,u=-2

Find the zeros of u^2-4u=12

The zeros of u^2-4u=12 are u=6,u=-2

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u^{2} - 4 u = 12

u = 6, u = - 2

Solution steps

u^{2} - 4 u = 12

Move

12

to the left side

u^{2} - 4 u - 12 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

u = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

u = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

u = 6, u = - 2

a, (a - b)^{2} = 12

x^{2} = 0.9

x, x^{2} + y^{2} + z^{2} - 4 y - 6 z + 13 = 0

4 x^{2} = - 17 x - 4

4.9 x^{(2)} - 5 x - 40 = 0

What are the solutions to the equation u^2-4u=12 ?
The solutions to the equation u^2-4u=12 are u=6,u=-2
Find the zeros of u^2-4u=12
The zeros of u^2-4u=12 are u=6,u=-2