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2-3x^2-5x=0
a = -3; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·(-3)·2
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*-3}=\frac{-2}{-6} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*-3}=\frac{12}{-6} =-2 $
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