3(3a-19)=(a-5)(a-3)

Solution for 3(3a-19)=(a-5)(a-3) equation:

3(3a-19)=(a-5)(a-3)

We move all terms to the left:

3(3a-19)-((a-5)(a-3))=0

We multiply parentheses

9a-((a-5)(a-3))-57=0

We multiply parentheses ..

-((+a^2-3a-5a+15))+9a-57=0


We calculate terms in parentheses: -((+a^2-3a-5a+15)), so:

(+a^2-3a-5a+15)

We get rid of parentheses

a^2-3a-5a+15

We add all the numbers together, and all the variables

a^2-8a+15

Back to the equation:

-(a^2-8a+15)


We add all the numbers together, and all the variables

9a-(a^2-8a+15)-57=0

We get rid of parentheses

-a^2+9a+8a-15-57=0

We add all the numbers together, and all the variables

-1a^2+17a-72=0

a = -1; b = 17; c = -72;
Δ = b2-4ac
Δ = 172-4·(-1)·(-72)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*-1}=\frac{-18}{-2}
=+9
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*-1}=\frac{-16}{-2}
=+8
$