4a^2+12a=135

Solution for 4a^2+12a=135 equation:

4a^2+12a=135

We move all terms to the left:

4a^2+12a-(135)=0

a = 4; b = 12; c = -135;
Δ = b2-4ac
Δ = 122-4·4·(-135)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-48}{2*4}=\frac{-60}{8}
=-7+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+48}{2*4}=\frac{36}{8}
=4+1/2
$