-3x^2+16x-1=0

Solution for -3x^2+16x-1=0 equation:

-3x^2+16x-1=0

a = -3; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·(-3)·(-1)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{61}}{2*-3}=\frac{-16-2\sqrt{61}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{61}}{2*-3}=\frac{-16+2\sqrt{61}}{-6}
$