a^2-15a+42=0

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Solution for a^2-15a+42=0 equation: 



a^2-15a+42=0

a = 1; b = -15; c = +42;
Δ = b2-4ac
Δ = -152-4·1·42
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{57}}{2*1}=\frac{15-\sqrt{57}}{2}
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{57}}{2*1}=\frac{15+\sqrt{57}}{2}
$

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