218=3x^2+21x-16

Solution for 218=3x^2+21x-16 equation:

218=3x^2+21x-16

We move all terms to the left:

218-(3x^2+21x-16)=0

We get rid of parentheses

-3x^2-21x+16+218=0

We add all the numbers together, and all the variables

-3x^2-21x+234=0

a = -3; b = -21; c = +234;
Δ = b2-4ac
Δ = -212-4·(-3)·234
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-57}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+57}{2*-3}=\frac{78}{-6}
=-13
$