3x^2+x=420

Solution for 3x^2+x=420 equation:

3x^2+x=420

We move all terms to the left:

3x^2+x-(420)=0

a = 3; b = 1; c = -420;
Δ = b2-4ac
Δ = 12-4·3·(-420)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-71}{2*3}=\frac{-72}{6}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+71}{2*3}=\frac{70}{6}
=11+2/3
$