16x^2+48x-92=0

Solution for 16x^2+48x-92=0 equation:

16x^2+48x-92=0

a = 16; b = 48; c = -92;
Δ = b2-4ac
Δ = 482-4·16·(-92)
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-64\sqrt{2}}{2*16}=\frac{-48-64\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+64\sqrt{2}}{2*16}=\frac{-48+64\sqrt{2}}{32}
$