Question

How do you solve `r^ { 2} - 10r - 19= 0`?

Answer 1

`r=(10\pmsqrt176)/2`

Explanation:

`r=(-b\pmsqrtb^2-4ac)/(2a)`
replace variables with numbers with a=1, b=-10, and c=-19
`r=-(-10)\pmsqrt((-10)^2-4(1)(-19))/(2(1))`
`r=(10\pmsqrt(100+76))/(2)`
`r=(10\pmsqrt176)/(2)`