5x2+21x+4=0

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Solution for 5x2+21x+4=0 equation:



5x^2+21x+4=0
a = 5; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·5·4
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*5}=\frac{-40}{10} =-4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*5}=\frac{-2}{10} =-1/5 $

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