2x+3x^2=161

Solution for 2x+3x^2=161 equation:

2x+3x^2=161

We move all terms to the left:

2x+3x^2-(161)=0

a = 3; b = 2; c = -161;
Δ = b2-4ac
Δ = 22-4·3·(-161)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-44}{2*3}=\frac{-46}{6}
=-7+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+44}{2*3}=\frac{42}{6}
=7
$