X^2+15x+42=0

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Solution for X^2+15x+42=0 equation: 



X^2+15X+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:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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

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