Question

How do you find the solution to the quadratic equation `x^2+8x+7=0`?

Answer 1

from the equation `ax^2+bx+c=0` you see that from your equation you have

a=1 b=8 c=7

`x=(-b\pm\sqrt(b^2-4ac))/(2a)`

put your numbers in the formula

`x=(-8\pm\sqrt(8^2-4*1*7))/(2*1)=(-8\pm\sqrt(64-28))/(2)`
`x=(-8\pm\sqrt(36))/(2)=(-8\pm6)/(2)=-4\pm3`

This give you two real solutions.

If inside the square root there was something negative, like `\sqrt(-7)` for an example at your level it has no solution, next years you'll learn about complex numbers and then in this group of numbers there is solution.

if inside the square root there was a 0 you would have only one solution right? `7\pm0=7`

so your equation has the answers x=-1 and x=-7 , you can see if it is right by putting the x inside your equation.

Never forget the signe of the constants a b and c and ALWAYS before start solving equale to 0, i mean, all the ehing different to zero have to be at the right or left of the equal signe, you can't start solving from `x^2 + 3x = 7` first you have to pass the seven to the left `x^2 + 3x-7 = 7-7=0`