4(12d-4)(d+3)=

Solution for 4(12d-4)(d+3)= equation:

4(12d-4)(d+3)=

We move all terms to the left:

4(12d-4)(d+3)-()=0

We add all the numbers together, and all the variables

4(12d-4)(d+3)=0

We multiply parentheses ..

4(+12d^2+36d-4d-12)=0

We multiply parentheses

48d^2+144d-16d-48=0

We add all the numbers together, and all the variables

48d^2+128d-48=0

a = 48; b = 128; c = -48;
Δ = b2-4ac
Δ = 1282-4·48·(-48)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-160}{2*48}=\frac{-288}{96}
=-3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+160}{2*48}=\frac{32}{96}
=1/3
$