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3x^2+6x-189=0
a = 3; b = 6; c = -189;
Δ = b2-4ac
Δ = 62-4·3·(-189)
Δ = 2304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-48}{2*3}=\frac{-54}{6} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+48}{2*3}=\frac{42}{6} =7 $
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