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

The solutions to the equation 2v^2+v-5=(v+1)^2 are v=3,v=-2

Find the zeros of 2v^2+v-5=(v+1)^2

The zeros of 2v^2+v-5=(v+1)^2 are v=3,v=-2

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2 v^{2} + v - 5 = (v + 1)^{2}

v = 3, v = - 2

Solution steps

2 v^{2} + v - 5 = (v + 1)^{2}

Expand

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

2 v^{2} + v - 5 = v^{2} + 2 v + 1

Move

1

to the left side

2 v^{2} + v - 6 = v^{2} + 2 v

Move

2 v

to the left side

2 v^{2} - v - 6 = v^{2}

Move

v^{2}

to the left side

v^{2} - v - 6 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

v = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

v = 3, v = - 2

425 = - 0.00121246 x^{2} + 876

3 2^{\frac{1}{3}} \cdot 2^{\frac{1}{3}}

- 2 (x - 1) - x + 3 = - 1

3 x \leq 9

4 y^{2} - 12 y + 9

What are the solutions to the equation 2v^2+v-5=(v+1)^2 ?
The solutions to the equation 2v^2+v-5=(v+1)^2 are v=3,v=-2
Find the zeros of 2v^2+v-5=(v+1)^2
The zeros of 2v^2+v-5=(v+1)^2 are v=3,v=-2