(2x^2)-19x=-39

Solution for (2x^2)-19x=-39 equation:

(2x^2)-19x=-39

We move all terms to the left:

(2x^2)-19x-(-39)=0

We add all the numbers together, and all the variables

2x^2-19x+39=0

a = 2; b = -19; c = +39;
Δ = b2-4ac
Δ = -192-4·2·39
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-7}{2*2}=\frac{12}{4}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+7}{2*2}=\frac{26}{4}
=6+1/2
$