k(k+3)=10-5

Solution for k(k+3)=10-5 equation:

k(k+3)=10-5

We move all terms to the left:

k(k+3)-(10-5)=0

We add all the numbers together, and all the variables

k(k+3)-5=0

We multiply parentheses

k^2+3k-5=0

a = 1; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·1·(-5)
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{29}}{2*1}=\frac{-3-\sqrt{29}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*1}=\frac{-3+\sqrt{29}}{2}
$