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3x^2-28x-11=0
a = 3; b = -28; c = -11;
Δ = b2-4ac
Δ = -282-4·3·(-11)
Δ = 916
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{916}=\sqrt{4*229}=\sqrt{4}*\sqrt{229}=2\sqrt{229}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{229}}{2*3}=\frac{28-2\sqrt{229}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{229}}{2*3}=\frac{28+2\sqrt{229}}{6} $
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