-3x^2+36x-1=16

Solution for -3x^2+36x-1=16 equation:

-3x^2+36x-1=16

We move all terms to the left:

-3x^2+36x-1-(16)=0

We add all the numbers together, and all the variables

-3x^2+36x-17=0

a = -3; b = 36; c = -17;
Δ = b2-4ac
Δ = 362-4·(-3)·(-17)
Δ = 1092
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:

$\sqrt{\Delta}=\sqrt{1092}=\sqrt{4*273}=\sqrt{4}*\sqrt{273}=2\sqrt{273}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{273}}{2*-3}=\frac{-36-2\sqrt{273}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{273}}{2*-3}=\frac{-36+2\sqrt{273}}{-6}
$