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7x^2=3+3
We move all terms to the left:
7x^2-(3+3)=0
We add all the numbers together, and all the variables
7x^2-6=0
a = 7; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·7·(-6)
Δ = 168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*7}=\frac{0-2\sqrt{42}}{14} =-\frac{2\sqrt{42}}{14} =-\frac{\sqrt{42}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*7}=\frac{0+2\sqrt{42}}{14} =\frac{2\sqrt{42}}{14} =\frac{\sqrt{42}}{7} $
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