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12k^2+20k-32=0
a = 12; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·12·(-32)
Δ = 1936
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{1936}=44$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-44}{2*12}=\frac{-64}{24} =-2+2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+44}{2*12}=\frac{24}{24} =1 $
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