19x=2x2+42

Solution for 19x=2x2+42 equation:

19x=2x^2+42

We move all terms to the left:

19x-(2x^2+42)=0

We get rid of parentheses

-2x^2+19x-42=0

a = -2; b = 19; c = -42;
Δ = b2-4ac
Δ = 192-4·(-2)·(-42)
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*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*-2}=\frac{-14}{-4}
=3+1/2
$