# Solve for k k(2k+5)=3 k(2k+5)=3
Simplify k(2k+5).
Simplify by multiplying through.
Apply the distributive property.
k(2k)+k⋅5=3
Reorder.
Rewrite using the commutative property of multiplication.
2k⋅k+k⋅5=3
Move 5 to the left of k.
2k⋅k+5⋅k=3
2k⋅k+5⋅k=3
2k⋅k+5⋅k=3
Multiply k by k by adding the exponents.
Move k.
2(k⋅k)+5⋅k=3
Multiply k by k.
2k2+5⋅k=3
2k2+5k=3
2k2+5k=3
Move 3 to the left side of the equation by subtracting it from both sides.
2k2+5k-3=0
Factor by grouping.
For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=2⋅-3=-6 and whose sum is b=5.
Factor 5 out of 5k.
2k2+5(k)-3=0
Rewrite 5 as -1 plus 6
2k2+(-1+6)k-3=0
Apply the distributive property.
2k2-1k+6k-3=0
2k2-1k+6k-3=0
Factor out the greatest common factor from each group.
Group the first two terms and the last two terms.
(2k2-1k)+6k-3=0
Factor out the greatest common factor (GCF) from each group.
k(2k-1)+3(2k-1)=0
k(2k-1)+3(2k-1)=0
Factor the polynomial by factoring out the greatest common factor, 2k-1.
(2k-1)(k+3)=0
(2k-1)(k+3)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
2k-1=0
k+3=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
2k-1=0
Add 1 to both sides of the equation.
2k=1
Divide each term by 2 and simplify.
Divide each term in 2k=1 by 2.
2k2=12
Cancel the common factor of 2.
Cancel the common factor.
2k2=12
Divide k by 1.
k=12
k=12
k=12
k=12
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
k+3=0
Subtract 3 from both sides of the equation.
k=-3
k=-3
The final solution is all the values that make (2k-1)(k+3)=0 true.
k=12,-3
The result can be shown in multiple forms.
Exact Form:
k=12,-3
Decimal Form:
k=0.5,-3
Solve for k k(2k+5)=3

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