4a^2+a-5=0

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Solution for 4a^2+a-5=0 equation: 



4a^2+a-5=0

a = 4; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·4·(-5)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*4}=\frac{-10}{8}
=-1+1/4
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*4}=\frac{8}{8}
=1
$

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