What are the solutions to the equation 4t^2+2t-(sqrt(3)+3)=0 ?

The solutions to the equation 4t^2+2t-(sqrt(3)+3)=0 are t=(sqrt(3))/2 ,t=-(1+sqrt(3))/2

Find the zeros of 4t^2+2t-(sqrt(3)+3)=0

The zeros of 4t^2+2t-(sqrt(3)+3)=0 are t=(sqrt(3))/2 ,t=-(1+sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 t^{2} + 2 t - (3 + 3) = 0

t = \frac{3}{2}, t = - \frac{1 + 3}{2}

Decimal Notation

t = 0.86602 \dots, t = - 1.36602 \dots

Solution steps

4 t^{2} + 2 t - (3 + 3) = 0

Expand

4 t^{2} + 2 t - (3 + 3) : 4 t^{2} + 2 t - 3 - 3

4 t^{2} + 2 t - 3 - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 ( - 3 - 3 )}{2 \cdot 4}

2^{2} - 4 \cdot 4 (- 3 - 3) = 43 + 2

t_{1, 2} = \frac{- 2 \pm ( 4 3 + 2 )}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 2 + 4 3 + 2}{2 \cdot 4}, t_{2} = \frac{- 2 - ( 4 3 + 2 )}{2 \cdot 4}

t = \frac{- 2 + 4 3 + 2}{2 \cdot 4} : \frac{3}{2}

t = \frac{- 2 - ( 4 3 + 2 )}{2 \cdot 4} : - \frac{1 + 3}{2}

The solutions to the quadratic equation are:

t = \frac{3}{2}, t = - \frac{1 + 3}{2}

x^{2} + 2 = - 4 x

c^{2} - 7 c + 3 = - 2 c

2^{2} + x^{2} = 1 0^{2}

x^{2} + 2 x + 30 = 0

0.15 x^{2} - 0.32 x - 3.6 = 0

What are the solutions to the equation 4t^2+2t-(sqrt(3)+3)=0 ?
The solutions to the equation 4t^2+2t-(sqrt(3)+3)=0 are t=(sqrt(3))/2 ,t=-(1+sqrt(3))/2
Find the zeros of 4t^2+2t-(sqrt(3)+3)=0
The zeros of 4t^2+2t-(sqrt(3)+3)=0 are t=(sqrt(3))/2 ,t=-(1+sqrt(3))/2