21y^2-5y-4=0

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Solution for 21y^2-5y-4=0 equation: 



21y^2-5y-4=0

a = 21; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·21·(-4)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*21}=\frac{-14}{42}
=-1/3
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*21}=\frac{24}{42}
=4/7
$

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