# How do you solve x^2-10x+25=35?

Oct 2, 2016

$x = 5 + \sqrt{35}$
or
$x = 5 - \sqrt{35}$

#### Explanation:

1. Take 35 from both sides so that the quadratic equation is equal to 0.
${x}^{2} - 10 x - 10 = 0$

2. Solve for x using the following equation:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(1\right) \left(- 10\right)}}{2 \left(1\right)}$
$= \frac{10 \pm \sqrt{100 + 40}}{2}$
$= \frac{10 \pm \sqrt{140}}{2}$
$= \frac{10 \pm \left(\sqrt{35} \cdot \sqrt{4}\right)}{2}$
$= \frac{2 \left(5\right) \pm 2 \left(\sqrt{35}\right)}{2}$
$= \frac{2 \left(5 \pm \sqrt{35}\right)}{2}$
$= 5 \pm \sqrt{35}$

Oct 2, 2016

$x = 5 \pm \sqrt{35}$

#### Explanation:

color(blue)(x^2-10x+25=35

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

$\rightarrow {x}^{2} - 10 x - 10 = 0$

This is a quadratic equation (in form of color(orange)(ax^2+bx+c=0)

We use the quadratic formula to solve it

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Where,

color(violet)(a=1

color(violet)(b=-10

color(violet)(c=-10

$\rightarrow x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(1\right) \left(- 10\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{10 \pm \sqrt{100 - \left(- 40\right)}}{2}$

$\rightarrow x = \frac{10 \pm \sqrt{100 + 40}}{2}$

$\rightarrow x = \frac{10 \pm \sqrt{140}}{2}$

$\rightarrow x = \frac{10 \pm \sqrt{4 \cdot 35}}{2}$

$\rightarrow x = \frac{10 \pm 2 \sqrt{35}}{2}$

$\rightarrow x = \frac{{\cancel{10}}^{5} \pm {\cancel{2}}^{1} \sqrt{35}}{{\cancel{2}}^{1}}$

rArrcolor(green)(x=5+-sqrt35

Oct 2, 2016

$x = 5 \pm \sqrt{35}$

#### Explanation:

Notice that the left hand side of the equation is already a perfect square trinomial.

So rather than reformulate, we can proceed as follows:

${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25 = 35$

Take the square root of both ends, allowing for both positive and negative square roots to get:

$x - 5 = \pm \sqrt{35}$

Add $5$ to both sides to find:

$x = 5 \pm \sqrt{35}$

Finally note that $35 = 5 \cdot 7$ has no square factors, so $\sqrt{35}$ is already in simplest form.