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6k^2+25k+25=0
a = 6; b = 25; c = +25;
Δ = b2-4ac
Δ = 252-4·6·25
Δ = 25
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{25}=5$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5}{2*6}=\frac{-30}{12} =-2+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5}{2*6}=\frac{-20}{12} =-1+2/3 $
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