What are the solutions to the equation -t^2-7t-4=-2t^2 ?

The solutions to the equation -t^2-7t-4=-2t^2 are t=(7+sqrt(65))/2 ,t=(7-sqrt(65))/2

Find the zeros of -t^2-7t-4=-2t^2

The zeros of -t^2-7t-4=-2t^2 are t=(7+sqrt(65))/2 ,t=(7-sqrt(65))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- t^{2} - 7 t - 4 = - 2 t^{2}

t = \frac{7 + 65}{2}, t = \frac{7 - 65}{2}

Decimal Notation

t = 7.53112 \dots, t = - 0.53112 \dots

Solution steps

- t^{2} - 7 t - 4 = - 2 t^{2}

Move

2 t^{2}

to the left side

t^{2} - 7 t - 4 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 1 \cdot (- 4) = 65

t_{1, 2} = \frac{- ( - 7 ) \pm 65}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 7 ) + 65}{2 \cdot 1}, t_{2} = \frac{- ( - 7 ) - 65}{2 \cdot 1}

t = \frac{- ( - 7 ) + 65}{2 \cdot 1} : \frac{7 + 65}{2}

t = \frac{- ( - 7 ) - 65}{2 \cdot 1} : \frac{7 - 65}{2}

The solutions to the quadratic equation are:

t = \frac{7 + 65}{2}, t = \frac{7 - 65}{2}

x, (x + 3)^{2} = 8 (y - 1)^{3}

x^{2} - 7 x + 12 = 20 x^{2} + x + 1

4 x^{2} + 12 x - 44 = 0

21 = 7 x^{2}

x, x^{2} - x y + y^{3} = 4

What are the solutions to the equation -t^2-7t-4=-2t^2 ?
The solutions to the equation -t^2-7t-4=-2t^2 are t=(7+sqrt(65))/2 ,t=(7-sqrt(65))/2
Find the zeros of -t^2-7t-4=-2t^2
The zeros of -t^2-7t-4=-2t^2 are t=(7+sqrt(65))/2 ,t=(7-sqrt(65))/2