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(3)=4F^2
We move all terms to the left:
(3)-(4F^2)=0
a = -4; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-4)·3
Δ = 48
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-4}=\frac{0-4\sqrt{3}}{-8} =-\frac{4\sqrt{3}}{-8} =-\frac{\sqrt{3}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-4}=\frac{0+4\sqrt{3}}{-8} =\frac{4\sqrt{3}}{-8} =\frac{\sqrt{3}}{-2} $
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