5x^2+x=20

Solution for 5x^2+x=20 equation:

5x^2+x=20

We move all terms to the left:

5x^2+x-(20)=0

a = 5; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·5·(-20)
Δ = 401
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{401}}{2*5}=\frac{-1-\sqrt{401}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{401}}{2*5}=\frac{-1+\sqrt{401}}{10}
$