21x^2+4=19x

Solution for 21x^2+4=19x equation:

21x^2+4=19x

We move all terms to the left:

21x^2+4-(19x)=0

a = 21; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·21·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*21}=\frac{14}{42}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*21}=\frac{24}{42}
=4/7
$