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6k^2+3=5k^2-4k
We move all terms to the left:
6k^2+3-(5k^2-4k)=0
We get rid of parentheses
6k^2-5k^2+4k+3=0
We add all the numbers together, and all the variables
k^2+4k+3=0
a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2} =-1 $
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