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(F)=-3F^2-15F-12
We move all terms to the left:
(F)-(-3F^2-15F-12)=0
We get rid of parentheses
3F^2+15F+F+12=0
We add all the numbers together, and all the variables
3F^2+16F+12=0
a = 3; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·3·12
Δ = 112
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{7}}{2*3}=\frac{-16-4\sqrt{7}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{7}}{2*3}=\frac{-16+4\sqrt{7}}{6} $
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