k^2-4(25)(9)=0

Solution for k^2-4(25)(9)=0 equation:

k^2-4(25)(9)=0

a = 1; b = 0; c = -4259;
Δ = b2-4ac
Δ = 02-4·1·(-4259)
Δ = 17036
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:

$\sqrt{\Delta}=\sqrt{17036}=\sqrt{4*4259}=\sqrt{4}*\sqrt{4259}=2\sqrt{4259}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{4259}}{2*1}=\frac{0-2\sqrt{4259}}{2}
=-\frac{2\sqrt{4259}}{2}
=-\sqrt{4259}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{4259}}{2*1}=\frac{0+2\sqrt{4259}}{2}
=\frac{2\sqrt{4259}}{2}
=\sqrt{4259}
$