What are the solutions to the equation 2v^2-13v+2=(v-4)^2 ?

The solutions to the equation 2v^2-13v+2=(v-4)^2 are v=7,v=-2

Find the zeros of 2v^2-13v+2=(v-4)^2

The zeros of 2v^2-13v+2=(v-4)^2 are v=7,v=-2

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2 v^{2} - 13 v + 2 = (v - 4)^{2}

v = 7, v = - 2

Solution steps

2 v^{2} - 13 v + 2 = (v - 4)^{2}

Expand

(v - 4)^{2} : v^{2} - 8 v + 16

2 v^{2} - 13 v + 2 = v^{2} - 8 v + 16

Move

16

to the left side

2 v^{2} - 13 v - 14 = v^{2} - 8 v

Move

8 v

to the left side

2 v^{2} - 5 v - 14 = v^{2}

Move

v^{2}

to the left side

v^{2} - 5 v - 14 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 14 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 14) = 9

v_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 5 ) + 9}{2 \cdot 1}, v_{2} = \frac{- ( - 5 ) - 9}{2 \cdot 1}

v = \frac{- ( - 5 ) + 9}{2 \cdot 1} : 7

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

The solutions to the quadratic equation are:

v = 7, v = - 2

(1 \frac{1}{3})^{3}

(4 x + 7) (x - 10) = 0

148 - 2 \leq x

8 = - \frac{2}{5} c

x - (1 - 3 x) = 8 x - 1

What are the solutions to the equation 2v^2-13v+2=(v-4)^2 ?
The solutions to the equation 2v^2-13v+2=(v-4)^2 are v=7,v=-2
Find the zeros of 2v^2-13v+2=(v-4)^2
The zeros of 2v^2-13v+2=(v-4)^2 are v=7,v=-2