4b^2-28b+48=0

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Solution for 4b^2-28b+48=0 equation: 



4b^2-28b+48=0

a = 4; b = -28; c = +48;
Δ = b2-4ac
Δ = -282-4·4·48
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*4}=\frac{24}{8}
=3
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*4}=\frac{32}{8}
=4
$

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