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12k^2+23k+5=0
a = 12; b = 23; c = +5;
Δ = b2-4ac
Δ = 232-4·12·5
Δ = 289
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{289}=17$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*12}=\frac{-40}{24} =-1+2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*12}=\frac{-6}{24} =-1/4 $
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