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3x^2+6x-1872=0
a = 3; b = 6; c = -1872;
Δ = b2-4ac
Δ = 62-4·3·(-1872)
Δ = 22500
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{22500}=150$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-150}{2*3}=\frac{-156}{6} =-26 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+150}{2*3}=\frac{144}{6} =24 $
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