# Solve for r 1r(r+1)=r+(r+1)+5 1r(r+1)=r+(r+1)+5
Simplify 1r(r+1).
Multiply r by 1.
r(r+1)=r+r+1+5
Apply the distributive property.
r⋅r+r⋅1=r+r+1+5
Simplify the expression.
Multiply r by r.
r2+r⋅1=r+r+1+5
Multiply r by 1.
r2+r=r+r+1+5
r2+r=r+r+1+5
r2+r=r+r+1+5
Simplify r+r+1+5.
Add r and r.
r2+r=2r+1+5
Add 1 and 5.
r2+r=2r+6
r2+r=2r+6
Move all terms containing r to the left side of the equation.
Subtract 2r from both sides of the equation.
r2+r-2r=6
Subtract 2r from r.
r2-r=6
r2-r=6
Move 6 to the left side of the equation by subtracting it from both sides.
r2-r-6=0
Factor r2-r-6 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 -6 and whose sum is -1.
-3,2
Write the factored form using these integers.
(r-3)(r+2)=0
(r-3)(r+2)=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-3=0
r+2=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
r-3=0
Add 3 to both sides of the equation.
r=3
r=3
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
r+2=0
Subtract 2 from both sides of the equation.
r=-2
r=-2
The final solution is all the values that make (r-3)(r+2)=0 true.
r=3,-2
Solve for r 1r(r+1)=r+(r+1)+5

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