16x^2+19x-20=0

Solution for 16x^2+19x-20=0 equation:

16x^2+19x-20=0

a = 16; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·16·(-20)
Δ = 1641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{1641}}{2*16}=\frac{-19-\sqrt{1641}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1641}}{2*16}=\frac{-19+\sqrt{1641}}{32}
$