Answer:
A quadratic equation in the form of [tex]ax^2+bx+c = 0[/tex]....[1], then the solution of the equation is given by:
[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
Given the equation:
[tex]x^2-10x+25=35[/tex]
Subtract 35 from both sides we have;
[tex]x^2-10x-10=0[/tex]
On comparing with [1] we have;
a = 1 , b = -10 and c = -10
then;
[tex]x = \frac{-(-10) \pm \sqrt{(-10)^2-4(1)(-10)}}{2(1)}[/tex]
⇒[tex]x = \frac{10 \pm \sqrt{100+40}}{2}[/tex]
⇒[tex]x = \frac{10 \pm \sqrt{140}}{2}[/tex]
Simplify:
[tex]x = \frac{10 \pm 2\sqrt{35}}{2}[/tex]
⇒[tex]x = 5 \pm \sqrt{35}[/tex]
Therefore, the value of x are:
[tex]x = 5+ \sqrt{35}[/tex] and [tex]x = 5-\sqrt{35}[/tex]
