16x^2-50x+9=0

Solution for 16x^2-50x+9=0 equation:

16x^2-50x+9=0

a = 16; b = -50; c = +9;
Δ = b2-4ac
Δ = -502-4·16·9
Δ = 1924
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:

$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{481}}{2*16}=\frac{50-2\sqrt{481}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{481}}{2*16}=\frac{50+2\sqrt{481}}{32}
$