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3x^2+5=103
We move all terms to the left:
3x^2+5-(103)=0
We add all the numbers together, and all the variables
3x^2-98=0
a = 3; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·3·(-98)
Δ = 1176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1176}=\sqrt{196*6}=\sqrt{196}*\sqrt{6}=14\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{6}}{2*3}=\frac{0-14\sqrt{6}}{6} =-\frac{14\sqrt{6}}{6} =-\frac{7\sqrt{6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{6}}{2*3}=\frac{0+14\sqrt{6}}{6} =\frac{14\sqrt{6}}{6} =\frac{7\sqrt{6}}{3} $
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