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-3x^2+16x-20=0
a = -3; b = 16; c = -20;
Δ = b2-4ac
Δ = 162-4·(-3)·(-20)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*-3}=\frac{-20}{-6} =3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*-3}=\frac{-12}{-6} =+2 $
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