-2f^2-3f+4=0

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Solution for -2f^2-3f+4=0 equation: 



-2f^2-3f+4=0

a = -2; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-2)·4
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*-2}=\frac{3-\sqrt{41}}{-4}
$


$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*-2}=\frac{3+\sqrt{41}}{-4}
$

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