4n^2+7n-3=0

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Solution for 4n^2+7n-3=0 equation: 



4n^2+7n-3=0

a = 4; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·4·(-3)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{97}}{2*4}=\frac{-7-\sqrt{97}}{8}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{97}}{2*4}=\frac{-7+\sqrt{97}}{8}
$

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