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3x^2+3=24x-42
We move all terms to the left:
3x^2+3-(24x-42)=0
We get rid of parentheses
3x^2-24x+42+3=0
We add all the numbers together, and all the variables
3x^2-24x+45=0
a = 3; b = -24; c = +45;
Δ = b2-4ac
Δ = -242-4·3·45
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6}{2*3}=\frac{18}{6} =3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6}{2*3}=\frac{30}{6} =5 $
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