X^2+3z-4=0

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Solution for X^2+3z-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{-8}{2}
=-4
$


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

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