Answer:
[tex] \large \boxed{(x=4,x=5)}[/tex]
Step-by-step explanation:
Here,
[tex]x^{2} -9x+20 = 0[/tex]
We can find the value of x by various methods.
Method 1By using formula:
[tex] \large \boxed{ x = \frac{ - b \pm \sqrt{ {b}^{2 } - 4ac } }{2a} }[/tex]
For [tex]ax²+bx+c=0[/tex]
Here,a = 1, b = 9, c = 20
Putting the values.
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ [/tex]
[tex]x = - \frac{ - ( - 9) \pm \sqrt{ {9}^{2} - 4 \times 1 \times 20} }{2 \times 1} \\ [/tex]
Taking positive one.
[tex]x=\frac{-(-9)+ \sqrt{81-80} }{2} \\ \\x=\frac{9+1}{2} \\ \\x=5[/tex]
Taking negative one.
[tex]x=\frac{-(-9)- \sqrt{81-80} }{2} \\ \\x=\frac{9-1}{2} \\ \\x=4[/tex]
x=5, x=4
[tex] \: [/tex]
Method 2
[tex]\implies x^{2} -9x+20=0\\ \\ \implies x^{2}-4x-5x+20=0\\ \\ \implies x(x-4)-5(x-4)=0\\ \\ \implies(x-5)(x-4)=0 [/tex]
x-5=0...I
x-4=0...II
Solving equation I
x-5=0
x=5
Solving equation II
x-4=0
x=4
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