u^2-14u+45=0

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Solution for u^2-14u+45=0 equation: 



u^2-14u+45=0

a = 1; b = -14; c = +45;
Δ = b2-4ac
Δ = -142-4·1·45
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*1}=\frac{10}{2}
=5
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*1}=\frac{18}{2}
=9
$

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