How do you evaluate #7x ^ { 2} - 5x + 6- ( 12x ^ { 2} + 9x - 8)#?

3 Answers
Feb 11, 2018

See below.

Explanation:

When we see a minus sign by itself without a number in front of the bracket, we need to multiply each term in the bracket by #-1#.

#therefore#

#-1*12x^2=-12x^2#
#-1*9x=-9x#
#-1*-8=8#

This means our expression is now

#7x^2-5x+6-12x^2-9x+8#
Now we simplify by combining like terms (#x^2,x# and constants).

#therefore#

Final answer : #-5x^2-14x+14#

Feb 11, 2018

see a step process below;

Explanation:

#7x^2 - 5x + 6 -(12x^2 + 9x - 8)#

Open the bracket and simplify..

#7x^2 - 5x + 6 - 12x^2 - 9x + 8#

#7x^2 - 12x^2 - 5x - 9x + 6 + 8#

#-5x^2 - 14x + 12 -> "Quadratic Equation"#

Now solving the quadratic equation..

Using;

#x = (-b +- sqrt (b^2 - 4ac))/(2a)#

Where;

#a = -5#

#b = -14#

#c = 12#

Substituting the values..

#x = (-(-14) +- sqrt ((-14)^2 - 4(-5)(12)))/(2(-5)#

#x = (14 +- sqrt (196 + 240))/(-10)#

#x = (14 +- sqrt436)/(-10)#

Hope this helps!

Feb 11, 2018

see a step process below;

Explanation:

#7x^2 - 5x + 6 -(12x^2 + 9x - 8)#

Open the bracket and simplify..

#7x^2 - 5x + 6 - 12x^2 - 9x + 8#

#7x^2 - 12x^2 - 5x - 9x + 6 + 8#

#-5x^2 - 14x + 12 -> "Quadratic Equation"#

Now solving the quadratic equation..

Using;

#x = (-b +- sqrt (b^2 - 4ac))/(2a)#

Where;

#a = -5#

#b = -14#

#c = 12#

Substituting the values..

#x = (-(-14) +- sqrt ((-14)^2 - 4(-5)(12)))/(2(-5)#

#x = (14 +- sqrt (196 + 240))/(-10)#

#x = (14 +- sqrt436)/(-10)#

Hope this helps!