What are the solutions to the equation 2n^2+4n=26 ?

The solutions to the equation 2n^2+4n=26 are n=-1+sqrt(14),n=-1-sqrt(14)

Find the zeros of 2n^2+4n=26

The zeros of 2n^2+4n=26 are n=-1+sqrt(14),n=-1-sqrt(14)

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2 n^{2} + 4 n = 26

n = - 1 + 14, n = - 1 - 14

Decimal Notation

n = 2.74165 \dots, n = - 4.74165 \dots

Solution steps

2 n^{2} + 4 n = 26

Move

26

to the left side

2 n^{2} + 4 n - 26 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 26 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 26) = 414

n_{1, 2} = \frac{- 4 \pm 4 14}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 4 + 4 14}{2 \cdot 2}, n_{2} = \frac{- 4 - 4 14}{2 \cdot 2}

n = \frac{- 4 + 4 14}{2 \cdot 2} : - 1 + 14

n = \frac{- 4 - 4 14}{2 \cdot 2} : - 1 - 14

The solutions to the quadratic equation are:

n = - 1 + 14, n = - 1 - 14

x^{2} + 4 x + 104 = 0

w^{2} + 5 w - 14 = 0

X^{2} + 16 X + 63 = 0

y, x^{2} + (y - 1)^{2} = 1

(1 + 3 x)^{2} = 5 x + 9

What are the solutions to the equation 2n^2+4n=26 ?
The solutions to the equation 2n^2+4n=26 are n=-1+sqrt(14),n=-1-sqrt(14)
Find the zeros of 2n^2+4n=26
The zeros of 2n^2+4n=26 are n=-1+sqrt(14),n=-1-sqrt(14)