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3x^2=450
We move all terms to the left:
3x^2-(450)=0
a = 3; b = 0; c = -450;
Δ = b2-4ac
Δ = 02-4·3·(-450)
Δ = 5400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5400}=\sqrt{900*6}=\sqrt{900}*\sqrt{6}=30\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{6}}{2*3}=\frac{0-30\sqrt{6}}{6} =-\frac{30\sqrt{6}}{6} =-5\sqrt{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{6}}{2*3}=\frac{0+30\sqrt{6}}{6} =\frac{30\sqrt{6}}{6} =5\sqrt{6} $
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