Question

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

Answer 1

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`

Answer 2

`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|`