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3(-x^2+6x-8)=0
We multiply parentheses
-3x^2+18x-24=0
a = -3; b = 18; c = -24;
Δ = b2-4ac
Δ = 182-4·(-3)·(-24)
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-3}=\frac{-24}{-6} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-3}=\frac{-12}{-6} =+2 $
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