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3x^2-195x=0
a = 3; b = -195; c = 0;
Δ = b2-4ac
Δ = -1952-4·3·0
Δ = 38025
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{38025}=195$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-195)-195}{2*3}=\frac{0}{6} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-195)+195}{2*3}=\frac{390}{6} =65 $
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