7a+(3a-4)(6-2a)=-6a+33a-24

Solution for 7a+(3a-4)(6-2a)=-6a+33a-24 equation:

7a+(3a-4)(6-2a)=-6a+33a-24

We move all terms to the left:

7a+(3a-4)(6-2a)-(-6a+33a-24)=0

We add all the numbers together, and all the variables

7a+(3a-4)(-2a+6)-(27a-24)=0

We get rid of parentheses

7a+(3a-4)(-2a+6)-27a+24=0

We multiply parentheses ..

(-6a^2+18a+8a-24)+7a-27a+24=0

We add all the numbers together, and all the variables

(-6a^2+18a+8a-24)-20a+24=0

We get rid of parentheses

-6a^2+18a+8a-20a-24+24=0

We add all the numbers together, and all the variables

-6a^2+6a=0

a = -6; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-6)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-6}=\frac{-12}{-12}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-6}=\frac{0}{-12}
=0
$