-2k^2+23k-20=0

Solution for -2k^2+23k-20=0 equation:

-2k^2+23k-20=0

a = -2; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·(-2)·(-20)
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3\sqrt{41}}{2*-2}=\frac{-23-3\sqrt{41}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3\sqrt{41}}{2*-2}=\frac{-23+3\sqrt{41}}{-4}
$