# Solve for v (v-4)^2=2v^2+2v+40

(v-4)2=2v2+2v+40
Since v is on the right side of the equation, switch the sides so it is on the left side of the equation.
2v2+2v+40=(v-4)2
Simplify (v-4)2.
Rewrite (v-4)2 as (v-4)(v-4).
2v2+2v+40=(v-4)(v-4)
Expand (v-4)(v-4) using the FOIL Method.
Apply the distributive property.
2v2+2v+40=v(v-4)-4(v-4)
Apply the distributive property.
2v2+2v+40=v⋅v+v⋅-4-4(v-4)
Apply the distributive property.
2v2+2v+40=v⋅v+v⋅-4-4v-4⋅-4
2v2+2v+40=v⋅v+v⋅-4-4v-4⋅-4
Simplify and combine like terms.
Simplify each term.
Multiply v by v.
2v2+2v+40=v2+v⋅-4-4v-4⋅-4
Move -4 to the left of v.
2v2+2v+40=v2-4⋅v-4v-4⋅-4
Multiply -4 by -4.
2v2+2v+40=v2-4v-4v+16
2v2+2v+40=v2-4v-4v+16
Subtract 4v from -4v.
2v2+2v+40=v2-8v+16
2v2+2v+40=v2-8v+16
2v2+2v+40=v2-8v+16
Move all terms containing v to the left side of the equation.
Subtract v2 from both sides of the equation.
2v2+2v+40-v2=-8v+16
Add 8v to both sides of the equation.
2v2+2v+40-v2+8v=16
Subtract v2 from 2v2.
v2+2v+40+8v=16
v2+10v+40=16
v2+10v+40=16
Move 16 to the left side of the equation by subtracting it from both sides.
v2+10v+40-16=0
Subtract 16 from 40.
v2+10v+24=0
Factor v2+10v+24 using the AC method.
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 24 and whose sum is 10.
4,6
Write the factored form using these integers.
(v+4)(v+6)=0
(v+4)(v+6)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
v+4=0
v+6=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
v+4=0
Subtract 4 from both sides of the equation.
v=-4
v=-4
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
v+6=0
Subtract 6 from both sides of the equation.
v=-6
v=-6
The final solution is all the values that make (v+4)(v+6)=0 true.
v=-4,-6
Solve for v (v-4)^2=2v^2+2v+40

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