What are the solutions to the equation (t+2)^2+(t-3)^2=9 ?

The solutions to the equation (t+2)^2+(t-3)^2=9 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2

Find the zeros of (t+2)^2+(t-3)^2=9

The zeros of (t+2)^2+(t-3)^2=9 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(t + 2)^{2} + (t - 3)^{2} = 9

t = \frac{1}{2} + i \frac{7}{2}, t = \frac{1}{2} - i \frac{7}{2}

Solution steps

(t + 2)^{2} + (t - 3)^{2} = 9

Expand

(t + 2)^{2} + (t - 3)^{2} : 2 t^{2} - 2 t + 13

2 t^{2} - 2 t + 13 = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 4 : 27 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = \frac{1}{2} + i \frac{7}{2}, t = \frac{1}{2} - i \frac{7}{2}

15 x^{2} + 21 x = 0

(x + 1) (x - 5) = 0

(x + 1)^{2} = 3

3 x^{2} - 5 = 13

6 x^{2} - 2 x + 3 = 0

What are the solutions to the equation (t+2)^2+(t-3)^2=9 ?
The solutions to the equation (t+2)^2+(t-3)^2=9 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2
Find the zeros of (t+2)^2+(t-3)^2=9
The zeros of (t+2)^2+(t-3)^2=9 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2