What are the solutions to the equation 4n^2-19n+18=0 ?

The solutions to the equation 4n^2-19n+18=0 are n=(19+sqrt(73))/8 ,n=(19-sqrt(73))/8

Find the zeros of 4n^2-19n+18=0

The zeros of 4n^2-19n+18=0 are n=(19+sqrt(73))/8 ,n=(19-sqrt(73))/8

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4 n^{2} - 19 n + 18 = 0

n = \frac{19 + 73}{8}, n = \frac{19 - 73}{8}

Decimal Notation

n = 3.44300 \dots, n = 1.30699 \dots

Solution steps

4 n^{2} - 19 n + 18 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 4 \cdot 18}{2 \cdot 4}

(- 19)^{2} - 4 \cdot 4 \cdot 18 = 73

n_{1, 2} = \frac{- ( - 19 ) \pm 73}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- ( - 19 ) + 73}{2 \cdot 4}, n_{2} = \frac{- ( - 19 ) - 73}{2 \cdot 4}

n = \frac{- ( - 19 ) + 73}{2 \cdot 4} : \frac{19 + 73}{8}

n = \frac{- ( - 19 ) - 73}{2 \cdot 4} : \frac{19 - 73}{8}

The solutions to the quadratic equation are:

n = \frac{19 + 73}{8}, n = \frac{19 - 73}{8}

1 - 4 x \leq 21

6 x^{2} + 6 < - 13 x + 4

\frac{1}{5} (x + 0.5) + 5.24 = \frac{3}{2} x + \frac{7}{10} (x + 2.2)

4 x^{2} + 32 x + 48

What are the solutions to the equation 4n^2-19n+18=0 ?
The solutions to the equation 4n^2-19n+18=0 are n=(19+sqrt(73))/8 ,n=(19-sqrt(73))/8
Find the zeros of 4n^2-19n+18=0
The zeros of 4n^2-19n+18=0 are n=(19+sqrt(73))/8 ,n=(19-sqrt(73))/8