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3x^2+102x=0
a = 3; b = 102; c = 0;
Δ = b2-4ac
Δ = 1022-4·3·0
Δ = 10404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{10404}=102$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-102}{2*3}=\frac{-204}{6} =-34 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+102}{2*3}=\frac{0}{6} =0 $
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