4j^2-11j+7=0

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Solution for 4j^2-11j+7=0 equation: 



4j^2-11j+7=0

a = 4; b = -11; c = +7;
Δ = b2-4ac
Δ = -112-4·4·7
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*4}=\frac{8}{8}
=1
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*4}=\frac{14}{8}
=1+3/4
$

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