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5+3x^2=55
We move all terms to the left:
5+3x^2-(55)=0
We add all the numbers together, and all the variables
3x^2-50=0
a = 3; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·3·(-50)
Δ = 600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{6}}{2*3}=\frac{0-10\sqrt{6}}{6} =-\frac{10\sqrt{6}}{6} =-\frac{5\sqrt{6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{6}}{2*3}=\frac{0+10\sqrt{6}}{6} =\frac{10\sqrt{6}}{6} =\frac{5\sqrt{6}}{3} $
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