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3x^2-3x=270
We move all terms to the left:
3x^2-3x-(270)=0
a = 3; b = -3; c = -270;
Δ = b2-4ac
Δ = -32-4·3·(-270)
Δ = 3249
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3249}=57$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-57}{2*3}=\frac{-54}{6} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+57}{2*3}=\frac{60}{6} =10 $
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