The solutions of the given quadratic equation are 6 and 7.
The polynomial equation whose highest degree is two is called a quadratic equation. It is expressed in the form of:
[tex]ax^{2} + bx +c =0[/tex]
where x is the unknown variable and a, b and c are the constant terms.
The "solutions" to the Quadratic Equation are where it is equal to zero.
According to the given question
we have
A quadratic equation [tex]x^{2} -2x + 49 = 11x + 7[/tex]
the above equation can be written as
[tex]x^{2} -2x-11x+49-7=0[/tex]
[tex]x^{2} -13x+42=0\\x^{2} -7x-6x+42=0\\x(x-7)-6(x-7)=0\\(x-6)(x-7)=0[/tex]
Let f(x) = (x -6)(x -7)
at x = 6
f(6) =(6- 6)(6 - 7) = 0× (-1) =0
at x = -7
f(x) = (-7-6)(-7-7) = -13××(-14) = 182
at x = -6
f(x) = (-6 -6)(-6-7) =-12×-13 =156
at x = 7
f(x) = (7-6)(7-7) = 1×0 =0
at x = 11
f(x) =(11-6)(11-7) = 5×4 =20
at x = 2
f(x) = (2 - 6)( 2- 7) = (-4)(-5) = 20
Since, at x =6 and x = 7 we are getting f(x) =0.
Hence, 6 and 7 are the solutions of the given quadratic equation.
Learn more about the solutions of the quadratic equations here:
https://brainly.com/question/2263981
#SPJ2
