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(-5)=-2F^2-10F
We move all terms to the left:
(-5)-(-2F^2-10F)=0
We add all the numbers together, and all the variables
-(-2F^2-10F)-5=0
We get rid of parentheses
2F^2+10F-5=0
a = 2; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·2·(-5)
Δ = 140
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*2}=\frac{-10-2\sqrt{35}}{4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*2}=\frac{-10+2\sqrt{35}}{4} $
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