a(2a-4)+9;a=3

Solution for a(2a-4)+9;a=3 equation:

a(2a-4)+9a=3

We move all terms to the left:

a(2a-4)+9a-(3)=0

We add all the numbers together, and all the variables

9a+a(2a-4)-3=0

We multiply parentheses

2a^2+9a-4a-3=0

We add all the numbers together, and all the variables

2a^2+5a-3=0

a = 2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·2·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*2}=\frac{-12}{4}
=-3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*2}=\frac{2}{4}
=1/2
$