l^2-48l+495=0

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Solution for l^2-48l+495=0 equation: 



l^2-48l+495=0

a = 1; b = -48; c = +495;
Δ = b2-4ac
Δ = -482-4·1·495
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$


$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-18}{2*1}=\frac{30}{2}
=15
$


$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+18}{2*1}=\frac{66}{2}
=33
$

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