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0=-3x^2-42x-120
We move all terms to the left:
0-(-3x^2-42x-120)=0
We add all the numbers together, and all the variables
-(-3x^2-42x-120)=0
We get rid of parentheses
3x^2+42x+120=0
a = 3; b = 42; c = +120;
Δ = b2-4ac
Δ = 422-4·3·120
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-18}{2*3}=\frac{-60}{6} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+18}{2*3}=\frac{-24}{6} =-4 $
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