x^2+15x+43=0

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Solution for x^2+15x+43=0 equation: 



x^2+15x+43=0

a = 1; b = 15; c = +43;
Δ = b2-4ac
Δ = 152-4·1·43
Δ = 53
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{53}}{2*1}=\frac{-15-\sqrt{53}}{2}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{53}}{2*1}=\frac{-15+\sqrt{53}}{2}
$

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