How do you solve #8x-10=x^2-7x+3# using the quadratic formula?

2 Answers
Aug 24, 2017

Answer:

Convert the given equation into standard quadratic form and then apply the quadratic formula to get
#color(white)("XXX")x=(15+-sqrt(173))/2#

Explanation:

The standard quadratic form is
#color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0#
with solutions given by the quadratic formula
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(green)c))/(2color(red)a)#

Converting the given equation:
#color(white)("XXX")8x-10=x^2-7x+3#
into standard form results in
#color(white)("XXX")color(red)1x^2+color(blue)(""(-15))x+color(green)(13)=0#

Plugging the coefficient values into the quadratic formula gives the result shown in the answer above.

Aug 24, 2017

Answer:

Answers(to 3 d.p.): 0.924 and 14.077

Explanation:

Since you have

#8x - 10 = ##x^2 - 7x+3#,

You transfer the numbers on the left hand side to the right in order to get the normal order: #ax^2 ± bx ± c#.

#8x - 10 = ##x^2 - 7x +3#

If you add 10 on both sides, you eliminate the -10 on the left and bring it to the right.

#8x - 10 +10 = ##x^2 - 7x +3 + 10#

= #8x = ##x^2 - 7x +13#

Do the same for the 8x until you get 0.

#8x -8x = ##x^2 - 7x +13-8x#

= #0 = ##x^2 - 15x +13#

= #x^2 - 15x +13#

So now that you have everything in order, you can input the coefficients into the quadratic formula:

#x=(-(-15) ± sqrt(225 - 52))/(2×1)#

#=(15 ± sqrt(173))/(2×1)#

#=(15 ± sqrt(173))/2#

#=(15 + sqrt(173))/2#

#=14.077#

#=(15 - sqrt(173))/2#

#=0.924#

Hope this helps!