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3x^2=20
We move all terms to the left:
3x^2-(20)=0
a = 3; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·3·(-20)
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*3}=\frac{0-4\sqrt{15}}{6} =-\frac{4\sqrt{15}}{6} =-\frac{2\sqrt{15}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*3}=\frac{0+4\sqrt{15}}{6} =\frac{4\sqrt{15}}{6} =\frac{2\sqrt{15}}{3} $
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