What are the solutions to the equation 4u^2-12u=0 ?

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

Find the zeros of 4u^2-12u=0

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

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4 u^{2} - 12 u = 0

u = 3, u = 0

Solution steps

4 u^{2} - 12 u = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 0 = 12

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

Separate the solutions

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

u = \frac{- ( - 12 ) + 12}{2 \cdot 4} : 3

u = \frac{- ( - 12 ) - 12}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

u = 3, u = 0

0 = - 2 x^{2} + 4 x + 16

4 x^{2} - 64 x + 76 = 0

x^{2} + 4 \cdot (- 12) = 0

2 u^{2} + 5 u - 3 = 0

2 x^{2} + 14 x = 16

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