d^2+7d+5=0

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Solution for d^2+7d+5=0 equation: 



d^2+7d+5=0

a = 1; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·1·5
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{29}}{2*1}=\frac{-7-\sqrt{29}}{2}
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{29}}{2*1}=\frac{-7+\sqrt{29}}{2}
$

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