In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

quadratic formula: x = startfraction negative b plus or minus startroot b squared minus 4 a c endroot over 2 a endfraction

x = 5plus or minus startroot 31 endroot
x = 5plus or minus startroot 19 endroot
x = 5plus or minus 2 startroot 19 endroot
x = 5plus or minus 2 startroot 31 endroot

x = 5 + √31 amd x = 5 - √31

Step-by-step explanation:

Rewrite 6 = x2 – 10x using " ^ " to denote exponentiation.  Then re-arrange the terms in descending powers of x:

x^2 – 10x - 6 = 0

Let's solve this by completing the square.  Add 25 to both sides and simplify the result:

x^2 - 10x + 25 = 6 + 25 = 31

The first three terms can be rewritten as

(x - 5)^2

and so we now have (x - 5)^2 = 31.

Taking the sqrt of both sides, we get

x - 5 = ± √31

And so the two real, unequal roots are

x = 5 + √31 amd x = 5 - √31

answer: 5000

step-by-step explanation:

marco saved more.   he saved $35 more.

step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35