What are the solutions to the equation (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) ?

The solutions to the equation (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) are u=1

Find the zeros of (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1)

The zeros of (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) are u=1

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(u + 2) (u + 1) + (u + 1) (3 u - 1) = (6 - u) (3 u - 1)

u = 1

Solution steps

(u + 2) (u + 1) + (u + 1) (3 u - 1) = (6 - u) (3 u - 1)

Expand

(u + 2) (u + 1) + (u + 1) (3 u - 1) : 4 u^{2} + 5 u + 1

Expand

(6 - u) (3 u - 1) : - 3 u^{2} + 19 u - 6

4 u^{2} + 5 u + 1 = - 3 u^{2} + 19 u - 6

Move

6

to the left side

4 u^{2} + 5 u + 7 = - 3 u^{2} + 19 u

Move

19 u

to the left side

4 u^{2} - 14 u + 7 = - 3 u^{2}

Move

3 u^{2}

to the left side

7 u^{2} - 14 u + 7 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 7 \cdot 7}{2 \cdot 7}

(- 14)^{2} - 4 \cdot 7 \cdot 7 = 0

u_{1, 2} = \frac{- ( - 14 ) \pm 0}{2 \cdot 7}

u = \frac{- ( - 14 )}{2 \cdot 7}

\frac{- ( - 14 )}{2 \cdot 7} = 1

u = 1

The solution to the quadratic equation is:

u = 1

x^{2} + x = - 6

10 n^{2} - 2 n = 0

15 x^{2} - 29 x - 14 = 0

5 = x^{2}

x^{2} + 9 x - 3 = 0

What are the solutions to the equation (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) ?
The solutions to the equation (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) are u=1
Find the zeros of (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1)
The zeros of (u+2)(u+1)+(u+1)(3u-1)=(6-u)(3u-1) are u=1