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28k^2+33k=0
a = 28; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·28·0
Δ = 1089
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{1089}=33$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*28}=\frac{-66}{56} =-1+5/28 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*28}=\frac{0}{56} =0 $
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