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3x^2+21x-132=0
a = 3; b = 21; c = -132;
Δ = b2-4ac
Δ = 212-4·3·(-132)
Δ = 2025
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{2025}=45$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-45}{2*3}=\frac{-66}{6} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+45}{2*3}=\frac{24}{6} =4 $
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