16x^2+40x+1.5=0

Solution for 16x^2+40x+1.5=0 equation:

16x^2+40x+1.5=0

a = 16; b = 40; c = +1.5;
Δ = b2-4ac
Δ = 402-4·16·1.5
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{94}}{2*16}=\frac{-40-4\sqrt{94}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{94}}{2*16}=\frac{-40+4\sqrt{94}}{32}
$