d^2+d-56=0

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Solution for d^2+d-56=0 equation: 



d^2+d-56=0

a = 1; b = 1; c = -56;
Δ = b2-4ac
Δ = 12-4·1·(-56)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$


$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*1}=\frac{-16}{2}
=-8
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*1}=\frac{14}{2}
=7
$

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