2x^2-3x-8=397

Solution for 2x^2-3x-8=397 equation:

2x^2-3x-8=397

We move all terms to the left:

2x^2-3x-8-(397)=0

We add all the numbers together, and all the variables

2x^2-3x-405=0

a = 2; b = -3; c = -405;
Δ = b2-4ac
Δ = -32-4·2·(-405)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-57}{2*2}=\frac{-54}{4}
=-13+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+57}{2*2}=\frac{60}{4}
=15
$