2x^2-9x-193=0

Solution for 2x^2-9x-193=0 equation:

2x^2-9x-193=0

a = 2; b = -9; c = -193;
Δ = b2-4ac
Δ = -92-4·2·(-193)
Δ = 1625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5\sqrt{65}}{2*2}=\frac{9-5\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5\sqrt{65}}{2*2}=\frac{9+5\sqrt{65}}{4}
$