b^2+7b-170=0

Solution for b^2+7b-170=0 equation:

b^2+7b-170=0

a = 1; b = 7; c = -170;
Δ = b2-4ac
Δ = 72-4·1·(-170)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-27}{2*1}=\frac{-34}{2}
=-17
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+27}{2*1}=\frac{20}{2}
=10
$