d^2+2d-48=0

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Solution for d^2+2d-48=0 equation: 



d^2+2d-48=0

a = 1; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$


$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*1}=\frac{-16}{2}
=-8
$


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

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