# Solve for r 3r(r+3)=30 3r(r+3)=30
Divide each term by 3 and simplify.
Divide each term in 3r(r+3)=30 by 3.
3r(r+3)3=303
Simplify 3r(r+3)3.
Cancel the common factor of 3.
Cancel the common factor.
3r(r+3)3=303
Divide r(r+3) by 1.
r(r+3)=303
r(r+3)=303
Apply the distributive property.
r⋅r+r⋅3=303
Simplify the expression.
Multiply r by r.
r2+r⋅3=303
Move 3 to the left of r.
r2+3r=303
r2+3r=303
r2+3r=303
Divide 30 by 3.
r2+3r=10
r2+3r=10
Move 10 to the left side of the equation by subtracting it from both sides.
r2+3r-10=0
Factor r2+3r-10 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 -10 and whose sum is 3.
-2,5
Write the factored form using these integers.
(r-2)(r+5)=0
(r-2)(r+5)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
r-2=0
r+5=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
r-2=0
Add 2 to both sides of the equation.
r=2
r=2
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
r+5=0
Subtract 5 from both sides of the equation.
r=-5
r=-5
The final solution is all the values that make (r-2)(r+5)=0 true.
r=2,-5
Solve for r 3r(r+3)=30

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