#1**0 **

For this, we need to transorm the eq. to standard form:

\(x^2-6x=\frac{1}{2}x+42\)

\(x^2-\frac{13}{2}x-42=0\)

Now to use the quadratic formula:

\(a=1\)

\(b=-\frac{13}{2}\)

\(c=-42\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-(-\frac{13}{2}) \pm \sqrt{(\frac{13}{2})^2-4(1)(-42)} \over 2(1)}\)

\(x = \frac{\frac{13}{2} \pm \sqrt{210+{1 \over 4}}}{2}\)

\(x = {{13 \over 2}\pm(14+{1 \over 2})\over 2}\)

\(x = {10+{1\over2}, -4}\)

Mathhemathh
Aug 16, 2017