How many real solutions does 2x^2-4x+3=5

Step-by-step explanation:

To find how many real solutions it has you have to find the solutions.

The quadratic formula is x=  

(the A with a weird line over it shouldn't be there but i cant get rid of it)

for the equation you gave me, you subtract 5 from both sides so it is equal to 0.  It should be:

This equation is now in the format:

Now plug in what you have for a, b, and c  (a= 2, b= -4, c= -2)

This will give you:

(again ignore that letter)

Now simplify it to:

There are 2 real solutions because 1 plus the square root of 2 and 1 minus the square root of two are both real. (If it was 1 plus or minus the square root of a negative number it would be 0 real solutions because you cant take the square root of a negative number)

Sorry if this is confusing

think about it deeply and passionately about this property

i am sorry

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°