-16x^2-26x+12=0

Solution for -16x^2-26x+12=0 equation:

-16x^2-26x+12=0

a = -16; b = -26; c = +12;
Δ = b2-4ac
Δ = -262-4·(-16)·12
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-38}{2*-16}=\frac{-12}{-32}
=3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+38}{2*-16}=\frac{64}{-32}
=-2
$