What are the solutions to the equation 11+19+27+(8n+3)=n(4n+7) ?

The solutions to the equation 11+19+27+(8n+3)=n(4n+7) are n=4,n=-15/4

Find the zeros of 11+19+27+(8n+3)=n(4n+7)

The zeros of 11+19+27+(8n+3)=n(4n+7) are n=4,n=-15/4

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11 + 19 + 27 + (8 n + 3) = n (4 n + 7)

n = 4, n = - \frac{15}{4}

Decimal Notation

n = 4, n = - 3.75

Solution steps

11 + 19 + 27 + (8 n + 3) = n (4 n + 7)

Expand

n (4 n + 7) : 4 n^{2} + 7 n

8 n + 60 = 4 n^{2} + 7 n

Switch sides

4 n^{2} + 7 n = 8 n + 60

Move

60

to the left side

4 n^{2} + 7 n - 60 = 8 n

Move

8 n

to the left side

4 n^{2} - n - 60 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 4 ( - 60 )}{2 \cdot 4}

(- 1)^{2} - 4 \cdot 4 (- 60) = 31

n_{1, 2} = \frac{- ( - 1 ) \pm 31}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 31}{2 \cdot 4}, n_{2} = \frac{- ( - 1 ) - 31}{2 \cdot 4}

n = \frac{- ( - 1 ) + 31}{2 \cdot 4} : 4

n = \frac{- ( - 1 ) - 31}{2 \cdot 4} : - \frac{15}{4}

The solutions to the quadratic equation are:

n = 4, n = - \frac{15}{4}

2 (x + 1)^{2} - 35 = 15

12 x^{2} = 24

2 (x + 8)^{2} - 19 = 31

6 - 6 x^{2} - 3 x = 0

12 x^{2} - 92 x + 120 = 0

What are the solutions to the equation 11+19+27+(8n+3)=n(4n+7) ?
The solutions to the equation 11+19+27+(8n+3)=n(4n+7) are n=4,n=-15/4
Find the zeros of 11+19+27+(8n+3)=n(4n+7)
The zeros of 11+19+27+(8n+3)=n(4n+7) are n=4,n=-15/4