16x^2+24x-272=0

Solution for 16x^2+24x-272=0 equation:

16x^2+24x-272=0

a = 16; b = 24; c = -272;
Δ = b2-4ac
Δ = 242-4·16·(-272)
Δ = 17984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17984}=\sqrt{64*281}=\sqrt{64}*\sqrt{281}=8\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{281}}{2*16}=\frac{-24-8\sqrt{281}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{281}}{2*16}=\frac{-24+8\sqrt{281}}{32}
$