We have the next given equation:
[tex]x^2+10x=3[/tex]A) Use the method of completing the square and square roots:
First, equal the equation to zero:
[tex]x^2+10x-3=0[/tex]Now, take half of the x term and square it:
[tex](10\ast\frac{1}{2})^2=25[/tex]Then add the result to both sides:
[tex]x^2+10x-3+25=25[/tex]Rewrite the perfect square on the left
[tex](x+5)^2=28[/tex]Now, take the square root of both sides:
[tex]\sqrt{(x+5)^2}=\pm\sqrt{28}[/tex]Simplify:
[tex]\begin{gathered} x+5=\pm\sqrt{4\ast7} \\ x+5=\pm2\sqrt{7} \end{gathered}[/tex]Finally, solve for x:
[tex]x=-5\pm2\sqrt{7}[/tex]B) Use the quadratic formula:
The quadratic formula is given by:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Use the form ax²+bx+c=0
Where
a=1
b=10
c=-3
Replacing:
[tex]\begin{gathered} x=\frac{-(10)\pm\sqrt{10^2-4(1)(-3)}}{2(1)} \\ x=\frac{-10\pm\sqrt{100+12}}{2} \\ \end{gathered}[/tex]Simplify the square root and then simplify the whole expression:
[tex]\begin{gathered} x=\frac{-10\pm\sqrt{16\ast7}}{2} \\ x=\frac{-10\pm4\sqrt{7}}{2} \\ x=\frac{2(-5\pm2\sqrt{7})}{2} \\ x=-5\pm2\sqrt{7} \end{gathered}[/tex]