What are the solutions to the equation -3n^2-n-10=-4n^2 ?

The solutions to the equation -3n^2-n-10=-4n^2 are n=(1+sqrt(41))/2 ,n=(1-sqrt(41))/2

Find the zeros of -3n^2-n-10=-4n^2

The zeros of -3n^2-n-10=-4n^2 are n=(1+sqrt(41))/2 ,n=(1-sqrt(41))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 n^{2} - n - 10 = - 4 n^{2}

n = \frac{1 + 41}{2}, n = \frac{1 - 41}{2}

Decimal Notation

n = 3.70156 \dots, n = - 2.70156 \dots

Solution steps

- 3 n^{2} - n - 10 = - 4 n^{2}

Move

4 n^{2}

to the left side

n^{2} - n - 10 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 10) = 41

n_{1, 2} = \frac{- ( - 1 ) \pm 41}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 41}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 41}{2 \cdot 1}

n = \frac{- ( - 1 ) + 41}{2 \cdot 1} : \frac{1 + 41}{2}

n = \frac{- ( - 1 ) - 41}{2 \cdot 1} : \frac{1 - 41}{2}

The solutions to the quadratic equation are:

n = \frac{1 + 41}{2}, n = \frac{1 - 41}{2}

(2 x - 1)^{2} + 4 = 0

x^{2} = 484

x^{2} = 576

8 x^{2} + 48 x + 72 = 0

x^{2} - 300 x + 22500 = 0

What are the solutions to the equation -3n^2-n-10=-4n^2 ?
The solutions to the equation -3n^2-n-10=-4n^2 are n=(1+sqrt(41))/2 ,n=(1-sqrt(41))/2
Find the zeros of -3n^2-n-10=-4n^2
The zeros of -3n^2-n-10=-4n^2 are n=(1+sqrt(41))/2 ,n=(1-sqrt(41))/2