4j^2+3j-28=0

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Solution for 4j^2+3j-28=0 equation: 



4j^2+3j-28=0

a = 4; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·4·(-28)
Δ = 457
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{457}}{2*4}=\frac{-3-\sqrt{457}}{8}
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{457}}{2*4}=\frac{-3+\sqrt{457}}{8}
$

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