16y^2+43y-48=0

Solution for 16y^2+43y-48=0 equation:

16y^2+43y-48=0

a = 16; b = 43; c = -48;
Δ = b2-4ac
Δ = 432-4·16·(-48)
Δ = 4921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-\sqrt{4921}}{2*16}=\frac{-43-\sqrt{4921}}{32}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{4921}}{2*16}=\frac{-43+\sqrt{4921}}{32}
$