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12u^2+34u=0
a = 12; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·12·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1156}=34$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*12}=\frac{-68}{24} =-2+5/6 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*12}=\frac{0}{24} =0 $
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