4a^2+12a=40

Solution for 4a^2+12a=40 equation:

4a^2+12a=40

We move all terms to the left:

4a^2+12a-(40)=0

a = 4; b = 12; c = -40;
Δ = b2-4ac
Δ = 122-4·4·(-40)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*4}=\frac{-40}{8}
=-5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*4}=\frac{16}{8}
=2
$