4a(a-3)=32

Solution for 4a(a-3)=32 equation:

4a(a-3)=32

We move all terms to the left:

4a(a-3)-(32)=0

We multiply parentheses

4a^2-12a-32=0

a = 4; b = -12; c = -32;
Δ = b2-4ac
Δ = -122-4·4·(-32)
Δ = 656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{41}}{2*4}=\frac{12-4\sqrt{41}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{41}}{2*4}=\frac{12+4\sqrt{41}}{8}
$