4s^2+4s-3=0

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



4s^2+4s-3=0

a = 4; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·4·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*4}=\frac{-12}{8}
=-1+1/2
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*4}=\frac{4}{8}
=1/2
$

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