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6k+3k(k+2)=5k+12
We move all terms to the left:
6k+3k(k+2)-(5k+12)=0
We multiply parentheses
3k^2+6k+6k-(5k+12)=0
We get rid of parentheses
3k^2+6k+6k-5k-12=0
We add all the numbers together, and all the variables
3k^2+7k-12=0
a = 3; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·3·(-12)
Δ = 193
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{-(7)-\sqrt{193}}{2*3}=\frac{-7-\sqrt{193}}{6} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{193}}{2*3}=\frac{-7+\sqrt{193}}{6} $
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