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54=3(4+x2)
We move all terms to the left:
54-(3(4+x2))=0
We add all the numbers together, and all the variables
-(3(+x^2+4))+54=0
We calculate terms in parentheses: -(3(+x^2+4)), so:We get rid of parentheses
3(+x^2+4)
We multiply parentheses
3x^2+12
Back to the equation:
-(3x^2+12)
-3x^2-12+54=0
We add all the numbers together, and all the variables
-3x^2+42=0
a = -3; b = 0; c = +42;
Δ = b2-4ac
Δ = 02-4·(-3)·42
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*-3}=\frac{0-6\sqrt{14}}{-6} =-\frac{6\sqrt{14}}{-6} =-\frac{\sqrt{14}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*-3}=\frac{0+6\sqrt{14}}{-6} =\frac{6\sqrt{14}}{-6} =\frac{\sqrt{14}}{-1} $
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