b2+7b-20=0

Solution for b2+7b-20=0 equation:

b2+7b-20=0

We add all the numbers together, and all the variables

b^2+7b-20=0

a = 1; b = 7; c = -20;
Δ = b2-4ac
Δ = 72-4·1·(-20)
Δ = 129
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{-(7)-\sqrt{129}}{2*1}=\frac{-7-\sqrt{129}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{129}}{2*1}=\frac{-7+\sqrt{129}}{2}
$