What are the solutions to the equation 2v^2-7v+21=(v+1)^2 ?

The solutions to the equation 2v^2-7v+21=(v+1)^2 are v=5,v=4

Find the zeros of 2v^2-7v+21=(v+1)^2

The zeros of 2v^2-7v+21=(v+1)^2 are v=5,v=4

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2 v^{2} - 7 v + 21 = (v + 1)^{2}

v = 5, v = 4

Solution steps

2 v^{2} - 7 v + 21 = (v + 1)^{2}

Expand

(v + 1)^{2} : v^{2} + 2 v + 1

2 v^{2} - 7 v + 21 = v^{2} + 2 v + 1

Move

1

to the left side

2 v^{2} - 7 v + 20 = v^{2} + 2 v

Move

2 v

to the left side

2 v^{2} - 9 v + 20 = v^{2}

Move

v^{2}

to the left side

v^{2} - 9 v + 20 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 20 = 1

v_{1, 2} = \frac{- ( - 9 ) \pm 1}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 9 ) + 1}{2 \cdot 1}, v_{2} = \frac{- ( - 9 ) - 1}{2 \cdot 1}

v = \frac{- ( - 9 ) + 1}{2 \cdot 1} : 5

v = \frac{- ( - 9 ) - 1}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

v = 5, v = 4

12 + y \leq 5

\frac{2}{5} - \frac{3}{8}

(\frac{1}{9})^{- \frac{3}{2}} \cdot (\frac{1}{8})^{\frac{2}{3}}

6 v^{2} - 12 v = 0

x^{2} + 8 x + 17 = 0

What are the solutions to the equation 2v^2-7v+21=(v+1)^2 ?
The solutions to the equation 2v^2-7v+21=(v+1)^2 are v=5,v=4
Find the zeros of 2v^2-7v+21=(v+1)^2
The zeros of 2v^2-7v+21=(v+1)^2 are v=5,v=4