Question

In simplest radical form, what are the solutions to the quadratic equation 6=x^2-10x?

Answer 1

Answer:

$5+2\sqrt{31$ and $5-2\sqrt{31$

Step-by-step explanation:

First move everything to the same side:

$6=x^2-10x$ ---> $x^2-10x-6=0$

Then use the quadratic formula:

x= (-b ± $\sqrt{b^2 -4ac}$)/2a

$a=1\\b=-10 \\c=-6$

x= (-(-10) ± $\sqrt{-10^2 -4*1*-6}$)/2*1

x= (10 ± $\sqrt{100+24}$)/2

Solution 1:

(10 + $\sqrt{1 24}$)/2

$\frac{10+2\sqrt{31} }{2}$= $5+2\sqrt{31$

Solution 1:

(10 - $\sqrt{1 24}$)/2

$\frac{10-2\sqrt{31} }{2}$= $5-2\sqrt{31$