-16a^2+24a+345=0

Solution for -16a^2+24a+345=0 equation:

-16a^2+24a+345=0

a = -16; b = 24; c = +345;
Δ = b2-4ac
Δ = 242-4·(-16)·345
Δ = 22656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{22656}=\sqrt{64*354}=\sqrt{64}*\sqrt{354}=8\sqrt{354}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{354}}{2*-16}=\frac{-24-8\sqrt{354}}{-32}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{354}}{2*-16}=\frac{-24+8\sqrt{354}}{-32}
$