16x^2+20x+2.5=0

Solution for 16x^2+20x+2.5=0 equation:

16x^2+20x+2.5=0

a = 16; b = 20; c = +2.5;
Δ = b2-4ac
Δ = 202-4·16·2.5
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*16}=\frac{-20-4\sqrt{15}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*16}=\frac{-20+4\sqrt{15}}{32}
$