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(3)=3F^2-15
We move all terms to the left:
(3)-(3F^2-15)=0
We get rid of parentheses
-3F^2+15+3=0
We add all the numbers together, and all the variables
-3F^2+18=0
a = -3; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-3)·18
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*-3}=\frac{0-6\sqrt{6}}{-6} =-\frac{6\sqrt{6}}{-6} =-\frac{\sqrt{6}}{-1} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*-3}=\frac{0+6\sqrt{6}}{-6} =\frac{6\sqrt{6}}{-6} =\frac{\sqrt{6}}{-1} $
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