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3x^2-20x-11=0
a = 3; b = -20; c = -11;
Δ = b2-4ac
Δ = -202-4·3·(-11)
Δ = 532
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{133}}{2*3}=\frac{20-2\sqrt{133}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{133}}{2*3}=\frac{20+2\sqrt{133}}{6} $
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