7-3b+b2=8

Solution for 7-3b+b2=8 equation:

7-3b+b2=8

We move all terms to the left:

7-3b+b2-(8)=0

We add all the numbers together, and all the variables

b^2-3b-1=0

a = 1; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·1·(-1)
Δ = 13
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{-(-3)-\sqrt{13}}{2*1}=\frac{3-\sqrt{13}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*1}=\frac{3+\sqrt{13}}{2}
$