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3x^2+18x=115
We move all terms to the left:
3x^2+18x-(115)=0
a = 3; b = 18; c = -115;
Δ = b2-4ac
Δ = 182-4·3·(-115)
Δ = 1704
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1704}=\sqrt{4*426}=\sqrt{4}*\sqrt{426}=2\sqrt{426}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{426}}{2*3}=\frac{-18-2\sqrt{426}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{426}}{2*3}=\frac{-18+2\sqrt{426}}{6} $
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