3a^2+12a-102=-3

Solution for 3a^2+12a-102=-3 equation:

3a^2+12a-102=-3

We move all terms to the left:

3a^2+12a-102-(-3)=0

We add all the numbers together, and all the variables

3a^2+12a-99=0

a = 3; b = 12; c = -99;
Δ = b2-4ac
Δ = 122-4·3·(-99)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{37}}{2*3}=\frac{-12-6\sqrt{37}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{37}}{2*3}=\frac{-12+6\sqrt{37}}{6}
$