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

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

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

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

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

n = 4, n = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 4, n = 3

2 x^{2} + 24 x - 6 = 0

1.25 \cdot 1 0^{- 9} x^{2} + 20 \cdot 1 0^{- 6} x + 0.125 = 0

52 + 2 q = 100 - q^{2}

9 x^{2} + 27 x + 14 = 0

1000 = \frac{1}{2} \cdot 0.47 \cdot \frac{π \cdot D ^{2}}{4} \cdot 1.2256 \cdot 5^{3}

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