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3x^2+24x+36=0
a = 3; b = 24; c = +36;
Δ = b2-4ac
Δ = 242-4·3·36
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12}{2*3}=\frac{-36}{6} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12}{2*3}=\frac{-12}{6} =-2 $
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