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3x^2-25x-12=0
a = 3; b = -25; c = -12;
Δ = b2-4ac
Δ = -252-4·3·(-12)
Δ = 769
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{-(-25)-\sqrt{769}}{2*3}=\frac{25-\sqrt{769}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{769}}{2*3}=\frac{25+\sqrt{769}}{6} $
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