16y2+12y-10=0

Solution for 16y2+12y-10=0 equation:

16y^2+12y-10=0

a = 16; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·16·(-10)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*16}=\frac{-40}{32}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*16}=\frac{16}{32}
=1/2
$