3x^2-18x=16

Solution for 3x^2-18x=16 equation:

3x^2-18x=16

We move all terms to the left:

3x^2-18x-(16)=0

a = 3; b = -18; c = -16;
Δ = b2-4ac
Δ = -182-4·3·(-16)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{129}}{2*3}=\frac{18-2\sqrt{129}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{129}}{2*3}=\frac{18+2\sqrt{129}}{6}
$