X^2+10=19x

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Solution for X^2+10=19x equation: 



X^2+10=19X

We move all terms to the left:

X^2+10-(19X)=0

a = 1; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·1·10
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{321}}{2*1}=\frac{19-\sqrt{321}}{2}
$


$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{321}}{2*1}=\frac{19+\sqrt{321}}{2}
$

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