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3x^2+10=307
We move all terms to the left:
3x^2+10-(307)=0
We add all the numbers together, and all the variables
3x^2-297=0
a = 3; b = 0; c = -297;
Δ = b2-4ac
Δ = 02-4·3·(-297)
Δ = 3564
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3564}=\sqrt{324*11}=\sqrt{324}*\sqrt{11}=18\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{11}}{2*3}=\frac{0-18\sqrt{11}}{6} =-\frac{18\sqrt{11}}{6} =-3\sqrt{11} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{11}}{2*3}=\frac{0+18\sqrt{11}}{6} =\frac{18\sqrt{11}}{6} =3\sqrt{11} $
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