solve

Using the quadratic formula, we get: \({-b \pm \sqrt{b^2-4ac} \over 2a}\Leftrightarrow\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^24\cdot \:1\cdot \:5}}{2\cdot \:1}\Leftrightarrow\frac{6\pm\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1} \Leftrightarrow\frac{6\pm\sqrt{16}}{2\cdot \:1}\Leftrightarrow\frac{6\pm4}{2}=\boxed{5}, \boxed{1}.\)

...or factoring (which you will eventually be able to do 'in your head')

(b-5)(b-1) = 0       so b-5 = 0 means b = 5       or   b-1 = 0     means b = 1

b^2-6b+5=0

Hello Guest!

\(\color{BrickRed}b^2-6b+5=0\)

       p         q

\(b=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ b=3\pm\sqrt{9-5}\\ b=3\pm2\)

\(b_1=5\\ b_2=1\)

  !