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(3x^2+5x-2)-(2x+18)=
We move all terms to the left:
(3x^2+5x-2)-(2x+18)-()=0
We add all the numbers together, and all the variables
(3x^2+5x-2)-(2x+18)=0
We get rid of parentheses
3x^2+5x-2x-2-18=0
We add all the numbers together, and all the variables
3x^2+3x-20=0
a = 3; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·3·(-20)
Δ = 249
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{249}}{2*3}=\frac{-3-\sqrt{249}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{249}}{2*3}=\frac{-3+\sqrt{249}}{6} $
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