16a^2+120a+144=0

Solution for 16a^2+120a+144=0 equation:

16a^2+120a+144=0

a = 16; b = 120; c = +144;
Δ = b2-4ac
Δ = 1202-4·16·144
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-72}{2*16}=\frac{-192}{32}
=-6
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+72}{2*16}=\frac{-48}{32}
=-1+1/2
$