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7+3x^2+4=25
We move all terms to the left:
7+3x^2+4-(25)=0
We add all the numbers together, and all the variables
3x^2-14=0
a = 3; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·3·(-14)
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*3}=\frac{0-2\sqrt{42}}{6} =-\frac{2\sqrt{42}}{6} =-\frac{\sqrt{42}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*3}=\frac{0+2\sqrt{42}}{6} =\frac{2\sqrt{42}}{6} =\frac{\sqrt{42}}{3} $
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