c^2-15c+54=0

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Solution for c^2-15c+54=0 equation: 



c^2-15c+54=0

a = 1; b = -15; c = +54;
Δ = b2-4ac
Δ = -152-4·1·54
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3}{2*1}=\frac{12}{2}
=6
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3}{2*1}=\frac{18}{2}
=9
$

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