b^2+7b+1=0

Solution for b^2+7b+1=0 equation:

b^2+7b+1=0

a = 1; b = 7; c = +1;
Δ = b2-4ac
Δ = 72-4·1·1
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{5}}{2*1}=\frac{-7-3\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{5}}{2*1}=\frac{-7+3\sqrt{5}}{2}
$