a^2+16-10a=0

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Solution for a^2+16-10a=0 equation: 



a^2+16-10a=0

a = 1; b = -10; c = +16;
Δ = b2-4ac
Δ = -102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*1}=\frac{4}{2}
=2
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*1}=\frac{16}{2}
=8
$

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