x^2-(3x)-4=0

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Solution for x^2-(3x)-4=0 equation: 



x^2-(3x)-4=0

a = 1; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{-2}{2}
=-1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{8}{2}
=4
$

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