How do you solve # 2c^2 - 7c = -5#?

2 Answers
Apr 5, 2016

Answer:

1 and 5/2

Explanation:

#y = 2c^2 - 7c + 5 = 0#.
Since a + b + c = 0, use shortcut. The 2 real rots are: 1 and #c/a = 5/2#

Apr 5, 2016

Answer:

#x=5/2,1#

Explanation:

#color(blue)(2c^2-7c=-5#

Add #5# both sides

#rarr2c^2-7c+5=-5+5#

#rarr2c^2-7c+5=0#

Now,this is a trinomial.

So,

We can solve this by factoring or using Quadratic equation

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

Factoring

#color(purple)(2c^2-7c+5=0#

Factor the equation

If you have any problem with factoring trinomials,Watch this video:

#rarr(2x-5)(x-1)=0#

Now we can say that

#1)color(orange)((x-1)=0#

#2)color(indigo)((2x-5)=0#

Solve for both of the equations

#1)color(orange)(x-1=0#

#color(green)(rArrx=1#

#2)color(indigo)(2x-5=0#

#rarr2x=5#

#color(green)(rArrx=5/2#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

Using Quadratic formula

#color(purple)(2c^2-7+5=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

Quadratic formula

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

Remember that #a,bandc# are the coefficients

So,

#color(violet)(a=2,b=-7,c=5#

#rarrx=(-(-7)+-sqrt(-7^2-4(2)(5)))/(2(2))#

#rarrx=(7+-sqrt(49-4(40)))/(4)#

#rarrx=(7+-sqrt(49-40))/(4)#

#rarrx=(7+-sqrt(9))/(4)#

#rarrx=(7+-3)/(4)#

Now we have two solutions

#1)color(orange)(x=(7+3)/(4)=10/4=5/2#

#2)color(indigo)(x=(7-3)/4=4/4=1#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

#:.color(blue)(ul bar |x=5/2,1|#