2b^2+59=b^2+17b-13

Solution for 2b^2+59=b^2+17b-13 equation:

2b^2+59=b^2+17b-13

We move all terms to the left:

2b^2+59-(b^2+17b-13)=0

We get rid of parentheses

2b^2-b^2-17b+13+59=0

We add all the numbers together, and all the variables

b^2-17b+72=0

a = 1; b = -17; c = +72;
Δ = b2-4ac
Δ = -172-4·1·72
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-1}{2*1}=\frac{16}{2}
=8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+1}{2*1}=\frac{18}{2}
=9
$