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-3x^2+10x-3=0
a = -3; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·(-3)·(-3)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*-3}=\frac{-18}{-6} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*-3}=\frac{-2}{-6} =1/3 $
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