16r^2+24r-9=0

Solution for 16r^2+24r-9=0 equation:

16r^2+24r-9=0

a = 16; b = 24; c = -9;
Δ = b2-4ac
Δ = 242-4·16·(-9)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{2}}{2*16}=\frac{-24-24\sqrt{2}}{32}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{2}}{2*16}=\frac{-24+24\sqrt{2}}{32}
$