What are the solutions to the equation 8n^2-4n=18 ?

The solutions to the equation 8n^2-4n=18 are n=(1+sqrt(37))/4 ,n=(1-sqrt(37))/4

Find the zeros of 8n^2-4n=18

The zeros of 8n^2-4n=18 are n=(1+sqrt(37))/4 ,n=(1-sqrt(37))/4

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8 n^{2} - 4 n = 18

n = \frac{1 + 37}{4}, n = \frac{1 - 37}{4}

Decimal

n = 1.77069 \dots, n = - 1.27069 \dots

Solution steps

8 n^{2} - 4 n = 18

Move

18

to the left side

8 n^{2} - 4 n - 18 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 8 ( - 18 )}{2 \cdot 8}

(- 4)^{2} - 4 \cdot 8 (- 18) = 437

n_{1, 2} = \frac{- ( - 4 ) \pm 4 37}{2 \cdot 8}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 4 37}{2 \cdot 8}, n_{2} = \frac{- ( - 4 ) - 4 37}{2 \cdot 8}

n = \frac{- ( - 4 ) + 4 37}{2 \cdot 8} : \frac{1 + 37}{4}

n = \frac{- ( - 4 ) - 4 37}{2 \cdot 8} : \frac{1 - 37}{4}

The solutions to the quadratic equation are:

n = \frac{1 + 37}{4}, n = \frac{1 - 37}{4}

9 x^{2} + 30 x + 25 = 0

2 - x + 4 = 5

2 (6 x + 3) = 3 (4 x + 3)

5 ∣ x ∣ = - 3

(3 + 8 i) + (7 - 2 i) - (6 - 5 i)

What are the solutions to the equation 8n^2-4n=18 ?
The solutions to the equation 8n^2-4n=18 are n=(1+sqrt(37))/4 ,n=(1-sqrt(37))/4
Find the zeros of 8n^2-4n=18
The zeros of 8n^2-4n=18 are n=(1+sqrt(37))/4 ,n=(1-sqrt(37))/4