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3x^2+2x-6.52=0
a = 3; b = 2; c = -6.52;
Δ = b2-4ac
Δ = 22-4·3·(-6.52)
Δ = 82.24
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-\sqrt{82.24}}{2*3}=\frac{-2-\sqrt{82.24}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{82.24}}{2*3}=\frac{-2+\sqrt{82.24}}{6} $
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