21x^2+79x-48=0

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Solution for 21x^2+79x-48=0 equation:



21x^2+79x-48=0
a = 21; b = 79; c = -48;
Δ = b2-4ac
Δ = 792-4·21·(-48)
Δ = 10273
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{-(79)-\sqrt{10273}}{2*21}=\frac{-79-\sqrt{10273}}{42} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(79)+\sqrt{10273}}{2*21}=\frac{-79+\sqrt{10273}}{42} $

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