3(q-5)=2(q+5)q=25

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Solution for 3(q-5)=2(q+5)q=25 equation: 



3(q-5)=2(q+5)q=25

We move all terms to the left:

3(q-5)-(2(q+5)q)=0

We multiply parentheses

3q-(2(q+5)q)-15=0



We calculate terms in parentheses: -(2(q+5)q), so:

2(q+5)q

We multiply parentheses

2q^2+10q

Back to the equation:

-(2q^2+10q)



We get rid of parentheses

-2q^2+3q-10q-15=0

We add all the numbers together, and all the variables

-2q^2-7q-15=0

a = -2; b = -7; c = -15;
Δ = b2-4ac
Δ = -72-4·(-2)·(-15)
Δ = -71
Delta is less than zero, so there is no solution for the equation

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