What are the solutions to the equation 1.1t^2+0.2t-63=0 ?

The solutions to the equation 1.1t^2+0.2t-63=0 are t=(-1+sqrt(6931))/(11),t=-(1+sqrt(6931))/(11)

Find the zeros of 1.1t^2+0.2t-63=0

The zeros of 1.1t^2+0.2t-63=0 are t=(-1+sqrt(6931))/(11),t=-(1+sqrt(6931))/(11)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1.1 t^{2} + 0.2 t - 63 = 0

t = \frac{- 1 + 6931}{11}, t = - \frac{1 + 6931}{11}

Decimal Notation

t = 7.47751 \dots, t = - 7.65932 \dots

Solution steps

1.1 t^{2} + 0.2 t - 63 = 0

Multiply both sides by

10

11 t^{2} + 2 t - 630 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 11 ( - 630 )}{2 \cdot 11}

2^{2} - 4 \cdot 11 (- 630) = 26931

t_{1, 2} = \frac{- 2 \pm 2 6931}{2 \cdot 11}

Separate the solutions

t_{1} = \frac{- 2 + 2 6931}{2 \cdot 11}, t_{2} = \frac{- 2 - 2 6931}{2 \cdot 11}

t = \frac{- 2 + 2 6931}{2 \cdot 11} : \frac{- 1 + 6931}{11}

t = \frac{- 2 - 2 6931}{2 \cdot 11} : - \frac{1 + 6931}{11}

The solutions to the quadratic equation are:

t = \frac{- 1 + 6931}{11}, t = - \frac{1 + 6931}{11}

z^{3} - 2 z + 6 z - 12

x + 1 = 2 x

15 x - 38 = 27 x - 74

\frac{224}{5 !}

x^{8} - y^{4}

What are the solutions to the equation 1.1t^2+0.2t-63=0 ?
The solutions to the equation 1.1t^2+0.2t-63=0 are t=(-1+sqrt(6931))/(11),t=-(1+sqrt(6931))/(11)
Find the zeros of 1.1t^2+0.2t-63=0
The zeros of 1.1t^2+0.2t-63=0 are t=(-1+sqrt(6931))/(11),t=-(1+sqrt(6931))/(11)