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135=3x^2
We move all terms to the left:
135-(3x^2)=0
a = -3; b = 0; c = +135;
Δ = b2-4ac
Δ = 02-4·(-3)·135
Δ = 1620
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{5}}{2*-3}=\frac{0-18\sqrt{5}}{-6} =-\frac{18\sqrt{5}}{-6} =-\frac{3\sqrt{5}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{5}}{2*-3}=\frac{0+18\sqrt{5}}{-6} =\frac{18\sqrt{5}}{-6} =\frac{3\sqrt{5}}{-1} $
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