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4k^2+1k-12=0
We add all the numbers together, and all the variables
4k^2+k-12=0
a = 4; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·4·(-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{-(1)-\sqrt{193}}{2*4}=\frac{-1-\sqrt{193}}{8} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*4}=\frac{-1+\sqrt{193}}{8} $
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