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3k^2+10k-57=0
a = 3; b = 10; c = -57;
Δ = b2-4ac
Δ = 102-4·3·(-57)
Δ = 784
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{784}=28$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-28}{2*3}=\frac{-38}{6} =-6+1/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+28}{2*3}=\frac{18}{6} =3 $
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