2x^2+x=20

Solution for 2x^2+x=20 equation:

2x^2+x=20

We move all terms to the left:

2x^2+x-(20)=0

a = 2; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·2·(-20)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{161}}{2*2}=\frac{-1-\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*2}=\frac{-1+\sqrt{161}}{4}
$