(3x^2+4)=19x

Solution for (3x^2+4)=19x equation:

(3x^2+4)=19x

We move all terms to the left:

(3x^2+4)-(19x)=0

We add all the numbers together, and all the variables

-19x+(3x^2+4)=0

We get rid of parentheses

3x^2-19x+4=0

a = 3; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·3·4
Δ = 313
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{-(-19)-\sqrt{313}}{2*3}=\frac{19-\sqrt{313}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{313}}{2*3}=\frac{19+\sqrt{313}}{6}
$