Question

Which are the solutions of x2 = 19x + 1?

Answer 1

$x^2=19x+1 \\ x^2 -19x-1=0 \\ x= \frac{19б \sqrt{(-19)^2-4*1*(-1)} }{2*1}= \frac{19б \sqrt{361+4} }{2}= \frac{19б \sqrt{365} }{2}$

Answer 2

The quadratic function has a degree of 2. The solution to the given quadratic function is $x=\frac{19\pm \sqrt{365}}{2}$

The solution to a quadratic equation

The quadratic function has a degree of 2. Given the quadratic function expressed as:

x^2 = 19x + 1

This can also be written as:

x^2 - 19x - 1 = 0

Using the general formula expressed as:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

From the equation

a = 1

b = -19
c = -1

Substitute

$x=\frac{19\pm \sqrt{19^2-4(1)(-1)}}{2} \\x=\frac{19\pm \sqrt{365}}{2}$

Hence the solution to the given quadratic function is $x=\frac{19\pm \sqrt{365}}{2}$

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