[tex] x^{2} -5x+1=3[/tex]
The first thing you should do to solve is set the equation equal to zero. Subtract 3 from each side:
[tex]x^{2} -5x+1=3 \\ \\ x^{2} -5x-2=0[/tex]
The resulting equation doesn't seem to be factorable. To solve for x, use the quadratic formula:
[tex]x= \frac{-b \pm \sqrt{b^2-4ac} }{2a} \\ \\ x= \frac{-(-5) \pm \sqrt{(-5)^2-4(1 )(-2)} }{2(1)} \\ \\ x= \frac{5 \pm \sqrt{25-4(-2)} }{2} \\ \\ x= \frac{5 \pm \sqrt{25+8} }{2} \\ \\ x= \frac{5 \pm \sqrt{33} }{2} \\ \\ x \approx \frac{5 \pm 5.74}{2} \\ \\ \\ \\ x \approx \frac{5+5.74}{2} \\ \\ x \approx \frac{10.74}{2} \\ \\ x \approx 5.37 \\ \\ \\ \\ x \approx \frac{5-5.74}{2} \\ \\ x \approx \frac{-0.74}{2} \\ \\ x \approx -0.37[/tex]
To the nearest hundredths place, the solutions to the original equation are x = 5.37 and x = -0.37.
