(2a-4)+2(a-5)=-3a(a+1)

Solution for (2a-4)+2(a-5)=-3a(a+1) equation:

(2a-4)+2(a-5)=-3a(a+1)

We move all terms to the left:

(2a-4)+2(a-5)-(-3a(a+1))=0

We multiply parentheses

(2a-4)+2a-(-3a(a+1))-10=0

We get rid of parentheses

2a+2a-(-3a(a+1))-4-10=0


We calculate terms in parentheses: -(-3a(a+1)), so:

-3a(a+1)

We multiply parentheses

-3a^2-3a

Back to the equation:

-(-3a^2-3a)


We add all the numbers together, and all the variables

-(-3a^2-3a)+4a-14=0

We get rid of parentheses

3a^2+3a+4a-14=0

We add all the numbers together, and all the variables

3a^2+7a-14=0

a = 3; b = 7; c = -14;
Δ = b2-4ac
Δ = 72-4·3·(-14)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{217}}{2*3}=\frac{-7-\sqrt{217}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{217}}{2*3}=\frac{-7+\sqrt{217}}{6}
$