What are the solutions to the equation 5n^2-4n-10=15 ?

The solutions to the equation 5n^2-4n-10=15 are n=(2+sqrt(129))/5 ,n=(2-sqrt(129))/5

Find the zeros of 5n^2-4n-10=15

The zeros of 5n^2-4n-10=15 are n=(2+sqrt(129))/5 ,n=(2-sqrt(129))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 n^{2} - 4 n - 10 = 15

n = \frac{2 + 129}{5}, n = \frac{2 - 129}{5}

Decimal Notation

n = 2.67156 \dots, n = - 1.87156 \dots

Solution steps

5 n^{2} - 4 n - 10 = 15

Move

15

to the left side

5 n^{2} - 4 n - 25 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 25 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 25) = 2129

n_{1, 2} = \frac{- ( - 4 ) \pm 2 129}{2 \cdot 5}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 2 129}{2 \cdot 5}, n_{2} = \frac{- ( - 4 ) - 2 129}{2 \cdot 5}

n = \frac{- ( - 4 ) + 2 129}{2 \cdot 5} : \frac{2 + 129}{5}

n = \frac{- ( - 4 ) - 2 129}{2 \cdot 5} : \frac{2 - 129}{5}

The solutions to the quadratic equation are:

n = \frac{2 + 129}{5}, n = \frac{2 - 129}{5}

∣ x - 1 ∣ \leq 3

x^{2} > - 2 x

4 x^{2} - 1

3 x^{3} + 14 x^{2} - 5 x

- (8)^{0}

What are the solutions to the equation 5n^2-4n-10=15 ?
The solutions to the equation 5n^2-4n-10=15 are n=(2+sqrt(129))/5 ,n=(2-sqrt(129))/5
Find the zeros of 5n^2-4n-10=15
The zeros of 5n^2-4n-10=15 are n=(2+sqrt(129))/5 ,n=(2-sqrt(129))/5