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12k^2+49k=0
a = 12; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·12·0
Δ = 2401
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{2401}=49$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*12}=\frac{-98}{24} =-4+1/12 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*12}=\frac{0}{24} =0 $
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