6(3a+1)-30=3(2a-4)a=

Solution for 6(3a+1)-30=3(2a-4)a= equation:

6(3a+1)-30=3(2a-4)a=

We move all terms to the left:

6(3a+1)-30-(3(2a-4)a)=0

We multiply parentheses

18a-(3(2a-4)a)+6-30=0


We calculate terms in parentheses: -(3(2a-4)a), so:

3(2a-4)a

We multiply parentheses

6a^2-12a

Back to the equation:

-(6a^2-12a)


We add all the numbers together, and all the variables

18a-(6a^2-12a)-24=0

We get rid of parentheses

-6a^2+18a+12a-24=0

We add all the numbers together, and all the variables

-6a^2+30a-24=0

a = -6; b = 30; c = -24;
Δ = b2-4ac
Δ = 302-4·(-6)·(-24)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-18}{2*-6}=\frac{-48}{-12}
=+4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+18}{2*-6}=\frac{-12}{-12}
=1
$