2k^2=10k+16

Solution for 2k^2=10k+16 equation:

2k^2=10k+16

We move all terms to the left:

2k^2-(10k+16)=0

We get rid of parentheses

2k^2-10k-16=0

a = 2; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·2·(-16)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{57}}{2*2}=\frac{10-2\sqrt{57}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{57}}{2*2}=\frac{10+2\sqrt{57}}{4}
$