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3x+3x^2=180
We move all terms to the left:
3x+3x^2-(180)=0
a = 3; b = 3; c = -180;
Δ = b2-4ac
Δ = 32-4·3·(-180)
Δ = 2169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2169}=\sqrt{9*241}=\sqrt{9}*\sqrt{241}=3\sqrt{241}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{241}}{2*3}=\frac{-3-3\sqrt{241}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{241}}{2*3}=\frac{-3+3\sqrt{241}}{6} $
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