2b^2+59=b^2+17b-3

Solution for 2b^2+59=b^2+17b-3 equation:

2b^2+59=b^2+17b-3

We move all terms to the left:

2b^2+59-(b^2+17b-3)=0

We get rid of parentheses

2b^2-b^2-17b+3+59=0

We add all the numbers together, and all the variables

b^2-17b+62=0

a = 1; b = -17; c = +62;
Δ = b2-4ac
Δ = -172-4·1·62
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{41}}{2*1}=\frac{17-\sqrt{41}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{41}}{2*1}=\frac{17+\sqrt{41}}{2}
$