7b^2+26b=10-7b

Solution for 7b^2+26b=10-7b equation:

7b^2+26b=10-7b

We move all terms to the left:

7b^2+26b-(10-7b)=0

We add all the numbers together, and all the variables

7b^2+26b-(-7b+10)=0

We get rid of parentheses

7b^2+26b+7b-10=0

We add all the numbers together, and all the variables

7b^2+33b-10=0

a = 7; b = 33; c = -10;
Δ = b2-4ac
Δ = 332-4·7·(-10)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-37}{2*7}=\frac{-70}{14}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+37}{2*7}=\frac{4}{14}
=2/7
$