x^2-13x+42=0.

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Solution for x^2-13x+42=0. equation:



x^2-13x+42=0.
a = 1; b = -13; c = +42;
Δ = b2-4ac
Δ = -132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*1}=\frac{12}{2} =6 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*1}=\frac{14}{2} =7 $

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