What are the solutions to the equation 7n^2+4n=8 ?

The solutions to the equation 7n^2+4n=8 are n=(2(sqrt(15)-1))/7 ,n=-(2(1+sqrt(15)))/7

Find the zeros of 7n^2+4n=8

The zeros of 7n^2+4n=8 are n=(2(sqrt(15)-1))/7 ,n=-(2(1+sqrt(15)))/7

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7 n^{2} + 4 n = 8

n = \frac{2 ( 15 - 1 )}{7}, n = - \frac{2 ( 1 + 15 )}{7}

Decimal Notation

n = 0.82085 \dots, n = - 1.39228 \dots

Solution steps

7 n^{2} + 4 n = 8

Move

8

to the left side

7 n^{2} + 4 n - 8 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 7 ( - 8 )}{2 \cdot 7}

4^{2} - 4 \cdot 7 (- 8) = 415

n_{1, 2} = \frac{- 4 \pm 4 15}{2 \cdot 7}

Separate the solutions

n_{1} = \frac{- 4 + 4 15}{2 \cdot 7}, n_{2} = \frac{- 4 - 4 15}{2 \cdot 7}

n = \frac{- 4 + 4 15}{2 \cdot 7} : \frac{2 ( 15 - 1 )}{7}

n = \frac{- 4 - 4 15}{2 \cdot 7} : - \frac{2 ( 1 + 15 )}{7}

The solutions to the quadratic equation are:

n = \frac{2 ( 15 - 1 )}{7}, n = - \frac{2 ( 1 + 15 )}{7}

v^{2} - 4 v - 5 = 0

x^{2} + 3 x - 18 = 0

x, f (x + 1) = 2 x^{2} + 8

x^{2} - 18 x + 42 = 0

r^{2} = 81

What are the solutions to the equation 7n^2+4n=8 ?
The solutions to the equation 7n^2+4n=8 are n=(2(sqrt(15)-1))/7 ,n=-(2(1+sqrt(15)))/7
Find the zeros of 7n^2+4n=8
The zeros of 7n^2+4n=8 are n=(2(sqrt(15)-1))/7 ,n=-(2(1+sqrt(15)))/7