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7k^2+22k+3=0
a = 7; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·7·3
Δ = 400
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{400}=20$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*7}=\frac{-42}{14} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*7}=\frac{-2}{14} =-1/7 $
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