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3+5x^2+3=33
We move all terms to the left:
3+5x^2+3-(33)=0
We add all the numbers together, and all the variables
5x^2-27=0
a = 5; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·5·(-27)
Δ = 540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*5}=\frac{0-6\sqrt{15}}{10} =-\frac{6\sqrt{15}}{10} =-\frac{3\sqrt{15}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*5}=\frac{0+6\sqrt{15}}{10} =\frac{6\sqrt{15}}{10} =\frac{3\sqrt{15}}{5} $
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