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3x^2+93x-192=0
a = 3; b = 93; c = -192;
Δ = b2-4ac
Δ = 932-4·3·(-192)
Δ = 10953
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{10953}=\sqrt{9*1217}=\sqrt{9}*\sqrt{1217}=3\sqrt{1217}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(93)-3\sqrt{1217}}{2*3}=\frac{-93-3\sqrt{1217}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(93)+3\sqrt{1217}}{2*3}=\frac{-93+3\sqrt{1217}}{6} $
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