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10=3x^2
We move all terms to the left:
10-(3x^2)=0
a = -3; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-3)·10
Δ = 120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*-3}=\frac{0-2\sqrt{30}}{-6} =-\frac{2\sqrt{30}}{-6} =-\frac{\sqrt{30}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*-3}=\frac{0+2\sqrt{30}}{-6} =\frac{2\sqrt{30}}{-6} =\frac{\sqrt{30}}{-3} $
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