4x2-3x-4=0.

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Solution for 4x2-3x-4=0. equation: 



4x^2-3x-4=0.

a = 4; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·4·(-4)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{73}}{2*4}=\frac{3-\sqrt{73}}{8}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{73}}{2*4}=\frac{3+\sqrt{73}}{8}
$

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