b2+b-4=0

Solution for b2+b-4=0 equation:

b2+b-4=0

We add all the numbers together, and all the variables

b^2+b-4=0

a = 1; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·1·(-4)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*1}=\frac{-1-\sqrt{17}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*1}=\frac{-1+\sqrt{17}}{2}
$