10b^2+10=170

Solution for 10b^2+10=170 equation:

10b^2+10=170

We move all terms to the left:

10b^2+10-(170)=0

We add all the numbers together, and all the variables

10b^2-160=0

a = 10; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·10·(-160)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*10}=\frac{-80}{20}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*10}=\frac{80}{20}
=4
$