What are the solutions to the equation u^2-u-12=0 ?

The solutions to the equation u^2-u-12=0 are u=4,u=-3

Find the zeros of u^2-u-12=0

The zeros of u^2-u-12=0 are u=4,u=-3

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u^{2} - u - 12 = 0

u = 4, u = - 3

Solution steps

u^{2} - u - 12 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 1}

u = \frac{- ( - 1 ) + 7}{2 \cdot 1} : 4

u = \frac{- ( - 1 ) - 7}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

u = 4, u = - 3

- x^{2} + 4 x + 5

a^{2} x^{2} + ab x - 2 b^{2} = 0

x^{2} + 64 x = 0

x^{2} - 1 = 2 x

(2 x - 10)^{2} = (x - 6)^{2} + (x + 8)^{2}

What are the solutions to the equation u^2-u-12=0 ?
The solutions to the equation u^2-u-12=0 are u=4,u=-3
Find the zeros of u^2-u-12=0
The zeros of u^2-u-12=0 are u=4,u=-3