10w^2+29w-21=0

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Solution for 10w^2+29w-21=0 equation:



10w^2+29w-21=0
a = 10; b = 29; c = -21;
Δ = b2-4ac
Δ = 292-4·10·(-21)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*10}=\frac{-70}{20} =-3+1/2 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*10}=\frac{12}{20} =3/5 $

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