3x^2+20=19x

Solution for 3x^2+20=19x equation:

3x^2+20=19x

We move all terms to the left:

3x^2+20-(19x)=0

a = 3; b = -19; c = +20;
Δ = b2-4ac
Δ = -192-4·3·20
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*3}=\frac{8}{6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*3}=\frac{30}{6}
=5
$