What are the solutions to the equation u^2-7u+12=0 ?

The solutions to the equation u^2-7u+12=0 are u=4,u=3

Find the zeros of u^2-7u+12=0

The zeros of u^2-7u+12=0 are u=4,u=3

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u^{2} - 7 u + 12 = 0

u = 4, u = 3

Solution steps

u^{2} - 7 u + 12 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

u = 4, u = 3

2 \geq - 3 - 3 x \geq - 7

\frac{- 14 + 22 i}{- 2 - 4 i}

\frac{1}{3} x = 1

2 y^{3} - 8 y^{2} + y - 4

6 r^{2} + 3 r - 3 = 0

What are the solutions to the equation u^2-7u+12=0 ?
The solutions to the equation u^2-7u+12=0 are u=4,u=3
Find the zeros of u^2-7u+12=0
The zeros of u^2-7u+12=0 are u=4,u=3