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3x^2+180x=180
We move all terms to the left:
3x^2+180x-(180)=0
a = 3; b = 180; c = -180;
Δ = b2-4ac
Δ = 1802-4·3·(-180)
Δ = 34560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{34560}=\sqrt{2304*15}=\sqrt{2304}*\sqrt{15}=48\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-48\sqrt{15}}{2*3}=\frac{-180-48\sqrt{15}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+48\sqrt{15}}{2*3}=\frac{-180+48\sqrt{15}}{6} $
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