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3x^2+39x+36=0
a = 3; b = 39; c = +36;
Δ = b2-4ac
Δ = 392-4·3·36
Δ = 1089
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{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-33}{2*3}=\frac{-72}{6} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+33}{2*3}=\frac{-6}{6} =-1 $
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