16x^2+24x-352=0

Solution for 16x^2+24x-352=0 equation:

16x^2+24x-352=0

a = 16; b = 24; c = -352;
Δ = b2-4ac
Δ = 242-4·16·(-352)
Δ = 23104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23104}=152$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-152}{2*16}=\frac{-176}{32}
=-5+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+152}{2*16}=\frac{128}{32}
=4
$