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60+3x(1x+20)=180
We move all terms to the left:
60+3x(1x+20)-(180)=0
We add all the numbers together, and all the variables
3x(x+20)+60-180=0
We add all the numbers together, and all the variables
3x(x+20)-120=0
We multiply parentheses
3x^2+60x-120=0
a = 3; b = 60; c = -120;
Δ = b2-4ac
Δ = 602-4·3·(-120)
Δ = 5040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5040}=\sqrt{144*35}=\sqrt{144}*\sqrt{35}=12\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-12\sqrt{35}}{2*3}=\frac{-60-12\sqrt{35}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+12\sqrt{35}}{2*3}=\frac{-60+12\sqrt{35}}{6} $
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