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3x^2-5=190
We move all terms to the left:
3x^2-5-(190)=0
We add all the numbers together, and all the variables
3x^2-195=0
a = 3; b = 0; c = -195;
Δ = b2-4ac
Δ = 02-4·3·(-195)
Δ = 2340
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{65}}{2*3}=\frac{0-6\sqrt{65}}{6} =-\frac{6\sqrt{65}}{6} =-\sqrt{65} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{65}}{2*3}=\frac{0+6\sqrt{65}}{6} =\frac{6\sqrt{65}}{6} =\sqrt{65} $
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