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

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

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

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

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

Decimal Notation

t = 3.70156 \dots, t = - 2.70156 \dots

Solution steps

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

Move

4 t^{2}

to the left side

t^{2} - t - 10 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 4.9 x^{2} - 8 x + 30 = 0

2 x + 3 x^{2} = 0

3 + 2 z = 21 z^{2}

2 x = 14 - 3 x^{2}

4 x (x - 2) = 1

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