What are the solutions to the equation v^2-8v+12=0 ?

The solutions to the equation v^2-8v+12=0 are v=6,v=2

Find the zeros of v^2-8v+12=0

The zeros of v^2-8v+12=0 are v=6,v=2

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v^{2} - 8 v + 12 = 0

v = 6, v = 2

Solution steps

v^{2} - 8 v + 12 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

v = 6, v = 2

- 6 x^{2} + 48 x

- 2 s + 3 = 5 s + 24

9 x^{2} - 6 x + 1

∣ 5 x ∣ = 40

∣ 2 x - 8 ∣ < 0

What are the solutions to the equation v^2-8v+12=0 ?
The solutions to the equation v^2-8v+12=0 are v=6,v=2
Find the zeros of v^2-8v+12=0
The zeros of v^2-8v+12=0 are v=6,v=2