19=2x^2+3x=5

Solution for 19=2x^2+3x=5 equation:

19=2x^2+3x=5

We move all terms to the left:

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

We get rid of parentheses

-2x^2-3x+19=0

a = -2; b = -3; c = +19;
Δ = b2-4ac
Δ = -32-4·(-2)·19
Δ = 161
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{-(-3)-\sqrt{161}}{2*-2}=\frac{3-\sqrt{161}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{161}}{2*-2}=\frac{3+\sqrt{161}}{-4}
$