Question

x^2+6x,+7 =0
solve by quadratic formula
Please delete my id please​

Answer 1

Answer:

$\huge\star\underline\mathbb{HELLO}$

x^2+6x-7=0

=>x^2+2*3x+3^2-9-7=0

=>(x+3)^2-16=0

=>(x+3)^2-(4)^2=0

=>(x+3+4)(x+3-4)=0

=>(x+7)(x-1)=0

=>x=1,-7

Formula used

(x+y)^2=x^2+2xy+y^2 and x^2-y^2=(x+y)(x-y).

Here is how you can solve this question:

First, bring the constant term -7 to the other side.

Then add 1/2(coefficient of x)^2 to both the sides.

Simplify and get the answer.

Answer 2

$\mathfrak{\huge{\red{\underline{\underline{Answer}}}}}$

${x}^{2} + 6x - 7 = 0 \\ = > {x}^{2} + 7x - x - 7 = 0 \\ = > x(x + 7) - 1(x + 7) = 0 \\ = > (x + 7)(x - 1) = 0$

Either,

$x + 7 = 0 \\ = > x = - 7$

Or,

$x - 1 = 0 \\ = > x = 1$