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6k^2+53k+40=0
a = 6; b = 53; c = +40;
Δ = b2-4ac
Δ = 532-4·6·40
Δ = 1849
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{1849}=43$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-43}{2*6}=\frac{-96}{12} =-8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+43}{2*6}=\frac{-10}{12} =-5/6 $
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