What are the solutions to the equation 5(a-2)^2+3(a+1)=25 ?

The solutions to the equation 5(a-2)^2+3(a+1)=25 are a=(17+sqrt(329))/(10),a=(17-sqrt(329))/(10)

Find the zeros of 5(a-2)^2+3(a+1)=25

The zeros of 5(a-2)^2+3(a+1)=25 are a=(17+sqrt(329))/(10),a=(17-sqrt(329))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 (a - 2)^{2} + 3 (a + 1) = 25

a = \frac{17 + 329}{10}, a = \frac{17 - 329}{10}

Decimal Notation

a = 3.51383 \dots, a = - 0.11383 \dots

Solution steps

5 (a - 2)^{2} + 3 (a + 1) = 25

Expand

5 (a - 2)^{2} + 3 (a + 1) : 5 a^{2} - 17 a + 23

5 a^{2} - 17 a + 23 = 25

Move

25

to the left side

5 a^{2} - 17 a - 2 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

(- 17)^{2} - 4 \cdot 5 (- 2) = 329

a_{1, 2} = \frac{- ( - 17 ) \pm 329}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- ( - 17 ) + 329}{2 \cdot 5}, a_{2} = \frac{- ( - 17 ) - 329}{2 \cdot 5}

a = \frac{- ( - 17 ) + 329}{2 \cdot 5} : \frac{17 + 329}{10}

a = \frac{- ( - 17 ) - 329}{2 \cdot 5} : \frac{17 - 329}{10}

The solutions to the quadratic equation are:

a = \frac{17 + 329}{10}, a = \frac{17 - 329}{10}

- 16 t^{2} + 32 t = 0

(0.494 - x) (2.725 - x) - (0.281)^{2} = 0

3 x^{2} - 15 x - 2 = 0

q, q^{2} + 3 pq = 22

\frac{( x + 5 ) ^{2}}{144} - \frac{( x - 2 ) ^{2}}{25} = 1

What are the solutions to the equation 5(a-2)^2+3(a+1)=25 ?
The solutions to the equation 5(a-2)^2+3(a+1)=25 are a=(17+sqrt(329))/(10),a=(17-sqrt(329))/(10)
Find the zeros of 5(a-2)^2+3(a+1)=25
The zeros of 5(a-2)^2+3(a+1)=25 are a=(17+sqrt(329))/(10),a=(17-sqrt(329))/(10)