Question

Solve for x in the equation x^2-10x+25=35.

Answer 1

Answer:

$x_{1}=5+\sqrt{35}$

$x_{2}=5-\sqrt{35}$

Step-by-step explanation:

the equation is

$x^2-10x+25=35$

to solve we need to make the equation equal to zero, for that we substract 35 on each side:

$x^2-10x+25-35=0\\x^2-10x-10=0$

we have now an equation in the general form: $ax^2+bx+c=0$

where

$a=1$

$b=-10$

$c=-10$

and we find the values for x with the general formula:

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

substituting known values

$x=\frac{-(-10)+-\sqrt{(-10)^2-4(1)(-10)} }{2(1)}\\x=\frac{10+-\sqrt{100+40} }{2}\\x=\frac{10+-\sqrt{140} }{2}\\x=\frac{10+-\sqrt{4*35} }{2}\\x=\frac{10+-2\sqrt{35} }{2}\\x=5+-\sqrt{35}$

since we had a quadratic equation we will have 2 answers, one taking the positive sign and the other with the negative sign:

$x_{1}=5+\sqrt{35}$ ≈ 10.916

$x_{2}=5-\sqrt{35}$ ≈ -0.916

Answer 2

So, this is too vague but I'll solve for both quadratic formula and find the discriminant.

Quadratic formula: x=5+√35 OR x=5−√35
Finding the discriminant: 141