12b+(-35)=b2

Solution for 12b+(-35)=b2 equation:

12b+(-35)=b2

We move all terms to the left:

12b+(-35)-(b2)=0

We add all the numbers together, and all the variables

-1b^2+12b-35=0

a = -1; b = 12; c = -35;
Δ = b2-4ac
Δ = 122-4·(-1)·(-35)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*-1}=\frac{-14}{-2}
=+7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*-1}=\frac{-10}{-2}
=+5
$