32x^2+48x+15=0

Solution for 32x^2+48x+15=0 equation:

32x^2+48x+15=0

a = 32; b = 48; c = +15;
Δ = b2-4ac
Δ = 482-4·32·15
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{6}}{2*32}=\frac{-48-8\sqrt{6}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{6}}{2*32}=\frac{-48+8\sqrt{6}}{64}
$