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(x+40)(3x)=180
We move all terms to the left:
(x+40)(3x)-(180)=0
We multiply parentheses
3x^2+120x-180=0
a = 3; b = 120; c = -180;
Δ = b2-4ac
Δ = 1202-4·3·(-180)
Δ = 16560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{16560}=\sqrt{144*115}=\sqrt{144}*\sqrt{115}=12\sqrt{115}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-12\sqrt{115}}{2*3}=\frac{-120-12\sqrt{115}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+12\sqrt{115}}{2*3}=\frac{-120+12\sqrt{115}}{6} $
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